[LeetCode] 1302. Deepest Leaves Sum

Given the root of a binary tree, return the sum of values of its deepest leaves.

Example 1:

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

Example 2:

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

Constraints:

The number of nodes in the tree is in the range [1, 104].
1 <= Node.val <= 100

层数最深叶子节点的和。

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。

这道题的 tag 给的是 DFS 但是这里我给一个更容易理解的 BFS 的做法吧。题目要你计算的是所有的叶子节点的节点值的和。我们可以这样做，对于每一层节点，我们去计算当前层所有节点的和。如果有下一层的话，那么我们把 res 重置为 0 即可。

时间O(n)

空间O(n)

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     public int deepestLeavesSum(TreeNode root) {
18         int res = 0;
19         Queue<TreeNode> queue = new LinkedList<>();
20         queue.offer(root);
21         while (!queue.isEmpty()) {
22             int size = queue.size();
23             res = 0;
24             for (int i = 0; i < size; i++) {
25                 TreeNode cur = queue.poll();
26                 res += cur.val;
27                 if (cur.left != null) {
28                     queue.offer(cur.left);
29                 }
30                 if (cur.right != null) {
31                     queue.offer(cur.right);
32                 }
33             }
34         }
35         return res;
36     }
37 }

LeetCode 题目总结