原题链接在这里：

题目：

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.

题解：

减掉当前节点值，判断是不是叶子节点同时满足sum==0即可。

Time Complexity: O(n), worst case是每个node都跑了一次.

Space: O(logn), stack space.

AC Java:

1 /** 2  * Definition for a binary tree node. 3  * public class TreeNode { 4  *     int val; 5  *     TreeNode left; 6  *     TreeNode right; 7  *     TreeNode(int x) { val = x; } 8  * } 9  */10 public class Solution {11     public boolean hasPathSum(TreeNode root, int sum) {12         if(root == null){13             return false;14         }15         sum-=root.val;16         if(root.left == null && root.right == null && sum == 0){17             return true;18         }else{19             return hasPathSum(root.left,sum) || hasPathSum(root.right,sum);20         }21     }22 }

跟上, , .