标签:商业 etc problems binary pes turn 结果 problem png
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
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/deepest-leaves-sum
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
树的层序遍历,每一层都计算一下综合,直到最后一层,返回结果。用时间来换空间。
/** * 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) { Queue<TreeNode> a = new ArrayDeque<>(); Queue<TreeNode> b = new ArrayDeque<>(); a.add(root); while (!a.isEmpty() ) { int sum = 0; while (!a.isEmpty()) { TreeNode poll = a.poll(); if (poll.left != null) { b.add(poll.left); } if (poll.right != null) { b.add(poll.right); } sum += poll.val; } if (b.isEmpty()) { return sum; } a = b; b = new ArrayDeque<>(); } return 0; } }
标签:商业 etc problems binary pes turn 结果 problem png
原文地址:https://www.cnblogs.com/wangzaiguli/p/14780970.html