Path Sum 判断二叉树的和 DFS处理

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Path Sum

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.

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ //DFS处理class Solution {public:    bool hasPathSum(TreeNode *root, int sum) {                return dfs(root,sum,0);    }        bool dfs(TreeNode *node,int sum,int cur)    {        if(node==NULL)            return false;        if(node->left==NULL&&node->right==NULL)            return sum==cur+node->val;        return dfs(node->left,sum,cur+node->val)||dfs(node->right,sum,cur+node->val);    }};

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