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.

[Thoughts]

二叉树遍历。遍历过程中累加节点值，当到达任意叶节点的时候，进行判断。

[Code]

1:       bool hasPathSum(TreeNode *root, int sum) {  2:            return hasPathSum(root, 0, sum);      3:       }  4:       bool hasPathSum(TreeNode *root, int sum, int target) {  5:            if(root == NULL) return false;  6:            sum += root->val;  7:            if(root->left == NULL && root->right == NULL) //leaf  8:            {  9:                 if(sum == target)  10:                      return true;  11:                 else  12:                      return false;  13:            }  14:            return hasPathSum(root->left, sum, target)   15:                   || hasPathSum(root->right, sum, target);      16:       }