Path Sum

题目来源LeetCode
题目描述

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

这道题和我做的 Binary Tree Paths差不多，也是用dfs来遍历，最后将每一条路的总和存入vector中，最后再将样本与容器中的元素进行比较，代码如下：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ void dfs(TreeNode*root, vector<int>&sums, int sum){     if(root->left == NULL && root->right == NULL){         sums.push_back(sum);         return;     }     if(root->left != NULL){         dfs(root->left, sums,sum+root->left->val);     }     if(root->right != NULL){         dfs(root->right, sums,sum+root->right->val);     } }class Solution {public:    bool hasPathSum(TreeNode* root, int sum) {        vector<int>sums;        int s;        if(root == NULL) return false;        s = root->val;        dfs(root,sums,s);        for(int i = 0; i < sums.size(); i++){            if(sums[i] == sum)return true;        }        return false;    }};