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.

Note: A leaf is a node with no children.

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.

代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
bool hasPathSum(struct TreeNode* root, int sum) {
    if(root==NULL)
        return false;
    if(root->left==NULL&&root->right==NULL)
    {
        if(sum==root->val)
            return true;
        else
            return false;
    }
    return hasPathSum(root->left,sum - root->val)||hasPathSum(root->right,sum - root->val);
}