Path Sum

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本文介绍了一种通过深度优先搜索(DFS)遍历二叉树的方法，旨在判断是否存在从根节点到叶子节点的路径，使得该路径上的节点值之和等于给定的目标值。以LeetCode题目为例，详细阐述了算法思路及其实现过程。

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;
    }
};