Q3_subtree_of_another_tree.cpp

// Given the roots of two binary trees root and subRoot, return true if there is a subtree of root with the same structure and node values of subRoot and false otherwise.

// A subtree of a binary tree tree is a tree that consists of a node in tree and all of this node's descendants. The tree tree could also be considered as a subtree of itself.

// Example 1:
// Input: root = [3,4,5,1,2], subRoot = [4,1,2]
// Output: true

// Example 2:
// Input: root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2]
// Output: false

#include<bits/stdc++.h>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode() : val(0), left(nullptr), right(nullptr) {}
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
    TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};

class Solution {
public:
    bool isSame(TreeNode* root,TreeNode* subRoot){
        if(root == NULL && subRoot == NULL)
            return true;
        if(root == NULL || subRoot == NULL)
            return false;
        if(root->val != subRoot->val)
            return false;
        return isSame(root->left,subRoot->left) && isSame(root->right,subRoot->right);
    }
    bool isSubtree(TreeNode* root, TreeNode* subRoot) {
        if(root == NULL)
            return false;
        return isSame(root,subRoot) || isSubtree(root->left,subRoot) || isSubtree(root->right,subRoot);
    }
};

// Time Complexity  : O(mn) 
// Space Complexity : O(m + n)
// m = number of nodes in tree rooted at "root"
// n = number of nodes in subtree rooted at "subRoot"