112. Path Sum

Well, pretty straightforward DFS solution

1:  /**  
2:   * Definition for a binary tree node.  
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 hasPathSum(TreeNode* root, int sum) {  
13:      if (root == NULL) return false;  
14:      if (root->left == NULL && root->right == NULL) return sum == root->val;  
15:      return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);  
16:    }  
17:  };