原题地址:https://oj.leetcode.com/problems/path-sum-ii/ 题意: Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5…

Path Sum II Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example: Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1 return [ [5,4,11,2], [5,8,4,5] ] 这个问题需要返…

Path Sum II Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1 return [ [5,4,11,2], [5,8,4,5] ] SOLUTION…

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1 return [ [5,4,11,2], [5,8,4,5] ] 这道二叉树路径之和在之前的基础上又需要找…

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example:Given the below binary tree andsum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1 return [ [5,4,11,2], [5,8,4,5] ] 题意:给定一数,在树中找出所有路径和等于该…

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. 通过一个p指针,遍历 二叉树,并将每次的值 保存在 sum2 中 . 遇到右节点,将右节点+depth 保存在 temp中,当再次使用 该节点时,根据depth 将sum2中的长度削减成 depth-1 /** * Definition for a binary tree node. * st…

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example: Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1 return [ [5,4,11,2], [5,8,4,5] ] 给定一个二叉树和一个值.找出全部根到叶…

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example:Given the below binary tree and sum = 22, 5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1 return [ [5,4,11,2], [5,8,4,5] ] Hide Tags Tree Depth…

题意: 给一棵二叉树,每个叶子到根的路径之和为sum的,将所有可能的路径装进vector返回. 思路: 节点的值可能为负的.这样子就必须到了叶节点才能判断,而不能中途进行剪枝. /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL)…

1.类中递归调用函数需要加self # Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: # @param root, a tree node # @param sum, an integer # @return a boolean def hasPath…