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‎100.相同的树.cpp

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/*
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* @lc app=leetcode.cn id=100 lang=cpp
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*
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* [100] 相同的树
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*
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* https://leetcode-cn.com/problems/same-tree/description/
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*
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* algorithms
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* Easy (55.48%)
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* Likes: 249
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* Dislikes: 0
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* Total Accepted: 44.5K
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* Total Submissions: 80.2K
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* Testcase Example: '[1,2,3]\n[1,2,3]'
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*
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* 给定两个二叉树，编写一个函数来检验它们是否相同。
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*
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* 如果两个树在结构上相同，并且节点具有相同的值，则认为它们是相同的。
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*
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* 示例 1:
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*
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* 输入: 1 1
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* ⁠ / \ / \
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* ⁠ 2 3 2 3
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*
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* ⁠ [1,2,3], [1,2,3]
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*
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* 输出: true
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*
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* 示例 2:
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*
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* 输入: 1 1
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* ⁠ / \
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* ⁠ 2 2
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*
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* ⁠ [1,2], [1,null,2]
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*
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* 输出: false
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*
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*
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* 示例 3:
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*
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* 输入: 1 1
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* ⁠ / \ / \
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* ⁠ 2 1 1 2
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*
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* ⁠ [1,2,1], [1,1,2]
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*
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* 输出: false
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*
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*
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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bool isSameTree(TreeNode* p, TreeNode* q) {
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vector<int>pnums;
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vector<int>qnums;
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pnums = num(p, pnums);
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qnums = num(q, qnums);
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if (pnums.size() != qnums.size()) return false;
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else {
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for (int i = 0; i<pnums.size(); i++) {
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if (pnums[i] != qnums[i]) return false;
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}
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}
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return true;
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}
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vector<int> num(TreeNode* p, vector<int>& nums) {
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TreeNode* root = p;
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if (!root) {
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nums.push_back(-5);
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return nums;
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}
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else {
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nums.push_back(root->val);
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num(root->left, nums);
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num(root->right, nums);
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}
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return nums;
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}
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};
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// @lc code=end
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‎101.对称二叉树.cpp

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/*
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* @lc app=leetcode.cn id=101 lang=cpp
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*
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* [101] 对称二叉树
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*
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* https://leetcode-cn.com/problems/symmetric-tree/description/
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*
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* algorithms
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* Easy (48.91%)
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* Likes: 488
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* Dislikes: 0
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* Total Accepted: 66.7K
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* Total Submissions: 136.3K
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* Testcase Example: '[1,2,2,3,4,4,3]'
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*
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* 给定一个二叉树，检查它是否是镜像对称的。
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*
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* 例如，二叉树 [1,2,2,3,4,4,3] 是对称的。
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*
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* ⁠ 1
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* ⁠ / \
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* ⁠ 2 2
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* ⁠/ \ / \
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* 3 4 4 3
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*
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*
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* 但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的:
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*
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* ⁠ 1
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* ⁠ / \
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* ⁠ 2 2
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* ⁠ \ \
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* ⁠ 3 3
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*
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*
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* 说明:
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*
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* 如果你可以运用递归和迭代两种方法解决这个问题，会很加分。
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*
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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bool isSymmetric(TreeNode* root) {
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return ismerror(root, root);
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}
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bool ismerror(TreeNode* t1, TreeNode* t2){
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if(t1==NULL&&t2==NULL) return true;
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if(t1==NULL||t2==NULL) return false;
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return(t1->val==t2->val)&&ismerror(t1->left, t2->right)&&ismerror(t2->left,t1->right);
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}
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};
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// @lc code=end
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‎102.二叉树的层次遍历.cpp

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/*
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* @lc app=leetcode.cn id=102 lang=cpp
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*
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* [102] 二叉树的层次遍历
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*
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* https://leetcode-cn.com/problems/binary-tree-level-order-traversal/description/
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*
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* algorithms
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* Medium (59.24%)
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* Likes: 296
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* Dislikes: 0
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* Total Accepted: 51.8K
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* Total Submissions: 87.4K
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* Testcase Example: '[3,9,20,null,null,15,7]'
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*
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* 给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。
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*
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* 例如:
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* 给定二叉树: [3,9,20,null,null,15,7],
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*
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* ⁠ 3
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* ⁠ / \
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* ⁠ 9 20
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* ⁠ / \
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* ⁠ 15 7
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*
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*
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* 返回其层次遍历结果：
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*
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* [
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* ⁠ [3],
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* ⁠ [9,20],
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* ⁠ [15,7]
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* ]
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*
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*
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*/
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// @lc code=start
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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vector<vector<int>> levelOrder(TreeNode* root) {
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vector<vector<int>> ans;
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queue<TreeNode*> subtree;
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if(root){
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subtree.push(root);
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levelth(subtree, ans);
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}
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return ans;
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}
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void levelth(queue<TreeNode*> &subtree, vector<vector<int>> &ans){
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TreeNode *cur;
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while(!subtree.empty()){
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vector<int> level;
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int width = subtree.size();
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for(int i=0; i<width; i++){
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cur = subtree.front();
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level.push_back(cur->val);
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subtree.pop();
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if(cur->left) subtree.push(cur->left);
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if(cur->right) subtree.push(cur->right);
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}
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ans.push_back(level);
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}
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}
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};
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// @lc code=end
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‎105.construct-binary-tree-from-preorder-and-inorder-traversal.cpp

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/*
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* @lc app=leetcode id=105 lang=cpp
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*
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* [105] Construct Binary Tree from Preorder and Inorder Traversal
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*
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* https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/
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*
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* algorithms
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* Medium (43.01%)
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* Likes: 2043
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* Dislikes: 56
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* Total Accepted: 252.7K
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* Total Submissions: 587.5K
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* Testcase Example: '[3,9,20,15,7]\n[9,3,15,20,7]'
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*
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* Given preorder and inorder traversal of a tree, construct the binary tree.
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*
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* Note:
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* You may assume that duplicates do not exist in the tree.
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*
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* For example, given
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*
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*
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* preorder = [3,9,20,15,7]
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* inorder = [9,3,15,20,7]
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*
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* Return the following binary tree:
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*
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*
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* ⁠ 3
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* ⁠ / \
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* ⁠ 9 20
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* ⁠ / \
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* ⁠ 15 7
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*
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*/
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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int pre=0;
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TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
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return subtree(preorder, inorder, 0, preorder.size()-1);
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}
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TreeNode* subtree(vector<int>& preorder, vector<int>& inorder, int l, int r){
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if(l > r) return NULL;
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if(l == r) return new TreeNode(preorder[pre++]);
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TreeNode* root = new TreeNode(preorder[pre++]);
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int id = find(inorder.begin()+l, inorder.begin()+l+r, preorder[pre-1]) - inorder.begin();
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root->left = subtree(preorder, inorder, l, id-1);
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root->right = subtree(preorder, inorder, id+1, r);
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return root;
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}
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};
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‎105.从前序与中序遍历序列构造二叉树.cpp

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/*
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* @lc app=leetcode.cn id=105 lang=cpp
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*
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* [105] 从前序与中序遍历序列构造二叉树
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*
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* https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/
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*
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* algorithms
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* Medium (61.35%)
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* Likes: 234
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* Dislikes: 0
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* Total Accepted: 23.3K
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* Total Submissions: 38K
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* Testcase Example: '[3,9,20,15,7]\n[9,3,15,20,7]'
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*
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* 根据一棵树的前序遍历与中序遍历构造二叉树。
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*
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* 注意:
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* 你可以假设树中没有重复的元素。
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*
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* 例如，给出
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*
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* 前序遍历 preorder = [3,9,20,15,7]
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* 中序遍历 inorder = [9,3,15,20,7]
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*
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* 返回如下的二叉树：
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*
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* ⁠ 3
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* ⁠ / \
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* ⁠ 9 20
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* ⁠ / \
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* ⁠ 15 7
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*
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*/
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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* };
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*/
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class Solution {
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public:
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int pre=0;
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TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
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return subtree(preorder, inorder, 0, preorder.size()-1);
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}
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TreeNode* subtree(vector<int>& preorder, vector<int>& inorder, int l, int r){
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if(l > r) return NULL;
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if(l == r) return new TreeNode(preorder[pre++]);
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TreeNode* root = new TreeNode(preorder[pre++]);
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int id = find(inorder.begin()+l, inorder.begin()+l+r, preorder[pre-1]) - inorder.begin();
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root->left = subtree(preorder, inorder, l, id-1);
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root->right = subtree(preorder, inorder, id+1, r);
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return root;
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}
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};
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