# 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.

```
              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1
```

## 递归 :

这道求二叉树的路径需要用深度优先算法DFS的思想来遍历每一条完整的路径，也就是利用递归不停找子节点的左右子节点，而调用递归函数的参数只有当前节点和sum值。首先，如果输入的是一个空节点，则直接返回false，如果如果输入的只有一个根节点，则比较当前根节点的值和参数sum值是否相同，若相同，返回true，否则false。 这个条件也是递归的终止条件。下面我们就要开始递归了，由于函数的返回值是Ture/False，我们可以同时两个方向一起递归，中间用或||连接，只要有一个是True，整个结果就是True。递归左右节点时，这时候的sum值应该是原sum值减去当前节点的值.

```
# 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
        if root.val == sum and root.left is None and root.right is None:
            return True
        return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val)
```

## DFS:

```
# 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
        stack = [(root, sum)]

        while stack:
            node, currentSum = stack.pop()
            if not node.left and not node.right and node.val == currentSum:
                return True
            if node.right:
                stack.append((node.right, currentSum-node.val))
            if node.left:
                stack.append((node.left,currentSum-node.val))    
        return False
```
