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1302. 层数最深叶子节点的和

English Version

题目描述

给你一棵二叉树的根节点 root ，请你返回 层数最深的叶子节点的和 。

 

示例 1：

输入：root = [1,2,3,4,5,null,6,7,null,null,null,null,8]
输出：15

示例 2：

输入：root = [6,7,8,2,7,1,3,9,null,1,4,null,null,null,5]
输出：19

 

提示：

• 树中节点数目在范围 [1, 104] 之间。
• 1 <= Node.val <= 100

解法

广度优先搜索。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def deepestLeavesSum(self, root: TreeNode) -> int:
        q = deque([root])
        s = 0
        while q:
            n = len(q)
            s = 0
            for _ in range(n):
                node = q.popleft()
                s += node.val
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
        return s

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int deepestLeavesSum(TreeNode root) {
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        int s = 0;
        while (!q.isEmpty()) {
            int n = q.size();
            s = 0;
            while (n-- > 0) {
                TreeNode node = q.poll();
                s += node.val;
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
        }
        return s;
    }
}

C++

/**
 * Definition for a binary tree node.
 * 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:
    int deepestLeavesSum(TreeNode* root) {
        queue<TreeNode*> q;
        q.push(root);
        int s = 0;
        while (!q.empty()) {
            int n = q.size();
            s = 0;
            while (n--) {
                auto node = q.front();
                q.pop();
                s += node->val;
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
        }
        return s;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func deepestLeavesSum(root *TreeNode) int {
	q := []*TreeNode{}
	q = append(q, root)
	s := 0
	for len(q) != 0 {
		n := len(q)
		s = 0
		for i := 0; i < n; i++ {
			node := q[0]
			q = q[1:]
			s += node.Val
			if node.Left != nil {
				q = append(q, node.Left)
			}
			if node.Right != nil {
				q = append(q, node.Right)
			}
		}
	}
	return s
}

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