題目鏈接：https://leetcode.com/PRoblems/path-sum/?tab=Description

題目描述：

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.

              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.
思路一：
最直觀的思路就是用DFS遍歷，遍歷過程中將路徑上各節(jié)點(diǎn)的值加起來，看最后到葉子節(jié)點(diǎn)時(shí)，是否等于sum,代碼不夠簡潔
代碼：
/** * 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:    bool haspathSum(TreeNode* root, int sum) {        bool flag=0;        int tmp=0;        DFS(root,sum,tmp,flag);        return flag;    }    void DFS(TreeNode* root,int sum,int tmp,bool& flag)    {        if(root==NULL)            return;        tmp+=root->val;        if(tmp==sum&&root->left==NULL&&root->right==NULL)        {            flag=1;            return;        }        DFS(root->left,sum,tmp,flag);        DFS(root->right,sum,tmp,flag);    }};思路二：
不用另外寫一個(gè)函數(shù)，就用給出的函數(shù)進(jìn)行遞歸調(diào)用，用sum逐步減去遍歷到的節(jié)點(diǎn)，無論左子樹還是右子樹滿足條件都行，代碼很簡潔
代碼：
/** * 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:    bool hasPathSum(TreeNode* root, int sum) {        if(root==NULL)            return 0;        if(sum-root->val==0&&root->left==NULL&&root->right==NULL)            return 1;        return hasPathSum(root->left,sum-root->val)||hasPathSum(root->right,sum-root->val);    }};