Path Sum（二叉树路径和）

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

采用递归的方式，C：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
  * };
 */
bool dfs(struct TreeNode *node, int sum, int cur);
bool hasPathSum(struct TreeNode *root, int sum) {
    return dfs(root, sum, 0);
}

bool dfs(struct TreeNode *node, int sum, int cur){
    if(node == NULL) return false;
    if(node->left == NULL && node->right == NULL)
        return cur+node->val == sum;
    return(dfs(node->left, sum, cur+node->val) || dfs(node->right, sum, cur+node->val));
}

C++：

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     bool dfs(TreeNode *node, int sum, int curSum)
     {
         if (node == NULL)
             return false;
         
         if (node->left == NULL && node->right == NULL)
             return curSum + node->val == sum;
                
         return dfs(node->left, sum, curSum + node->val) || dfs(node->right, sum, curSum + node->val);
     }
     
     bool hasPathSum(TreeNode *root, int sum) {
         // Start typing your C/C++ solution below
         // DO NOT write int main() function
         return dfs(root, sum, 0);
     }
 };