[LeetCode] 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.

DFS或者BFS。同时，当本节点是叶子节点才判断和，否则节点为空时返回false

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     bool dfs(TreeNode *node, int sum, int curSum)
13     {
14         if (node == NULL)
15             return false;
16         
17         if (node->left == NULL && node->right == NULL)
18             return curSum + node->val == sum;
19                
20         return dfs(node->left, sum, curSum + node->val) || dfs(node->right, sum, curSum + node->val);
21     }
22     
23     bool hasPathSum(TreeNode *root, int sum) {
24         // Start typing your C/C++ solution below
25         // DO NOT write int main() function
26         return dfs(root, sum, 0);
27     }
28 };