## Path Sum I:

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.

### Analysis:

This is a basic practice for depth first search algorithm.

The way of thinking this problem:

• This question requires only true/false return, so the final result is the "or" operation of all the possible root-to-leaf path.
• Parameters used in recursion should only be the sum value and the current node.
• Condition:  if the sum == 0 and the node is leaf node return true,
•                         if the sum != 0 and the node is NULL, return false
•                         else return go root.left, sum-val || go root.right, sum-val
• How to check if the current node is a leaf node?  Both its left and right children all empty.

### Code(C++):

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool hasPathSum(TreeNode *root, int sum) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (root==NULL) {return false;}
if ( (root->left==NULL) && (root->right==NULL) && (sum-root->val==0) ) {return true;}
return (hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val));
}
};


### Code(Python):

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
def hasPathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: bool
"""
if not root:
return False
else:
sum = sum - root.val
if not root.left and not root.right:
if sum==0:
return True
return self.hasPathSum(root.left, sum) or self.hasPathSum(root.right, sum)