Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 这道题给我们一个二叉树,让我们返回所有根到叶节点的路径,跟之前那道Path Sum II 二叉树路径之和之二很类似,比那道稍微简单一…

Given a binary tree, return all root-to-leaf paths.Example Given the following binary tree: 1 /   \2     3 \  5 All root-to-leaf paths are: [  "1->2->5",  "1->3"] LeetCode上的原题,请参见我之前的博客Binary Tree Paths. 解法一: class Solution {…

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 给一个二叉树,返回所有根到叶节点的路径. Java: /** * Definition for a binary tree node…

Given a binary tree, return all root-to-leaf paths. Note: A leaf is a node with no children. Example: Input: 1 / \ 2 3 \ 5 Output: ["1->2->5", "1->3"] Explanation: All root-to-leaf paths are: 1->2->5, 1->3 题目 给定一棵二叉树,…

给定一个二叉树,返回从根节点到叶节点的所有路径.例如,给定以下二叉树:   1 /   \2     3 \  5所有根到叶路径是:["1->2->5", "1->3"] 详见:https://leetcode.com/problems/binary-tree-paths/description/ Java实现: /** * Definition for a binary tree node. * public class TreeNode { *…

找到所有根到叶子的路径 深度优先搜索(DFS), 即二叉树的先序遍历. /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<string> v…

http://blog.csdn.net/crazy1235/article/details/51474128 花样做二叉树的题……居然还是不会么…… /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ cl…

Given a binary tree, return all root-to-leaf paths. For example, given the following binary tree: 1 / \ 2 3 \ 5 All root-to-leaf paths are: ["1->2->5", "1->3"] 思路:用两个stack<TreeNode*> in , s; in : 记录当前的路径 p  , 和vector<…

题意: 给出一个二叉树,输出根到所有叶子节点的路径. 思路: 直接DFS一次,只需要判断是否到达了叶子,是就收集答案. /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { v…

Given a binary tree, return the tilt of the whole tree. The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0. The tilt of the …