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.

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

我们遍历一条路径结束时，判断这条路径所有节点的和是否等于sum，如果相等的话，我们返回true就可以了，用递归的解法如下.

class Solution {
  public:
    bool hasPathSum(TreeNode* root, int sum) {
        if(!root)
            return false;
        if(root->val == sum && root->right == NULL && root->left == NULL)
            return true;

        return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);
    }
};