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      1return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

 比较通俗易懂一点的看法：

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) return false;13         return helper(root, sum);14     }15     16     public boolean helper(TreeNode root, int remain) {17         if (root.left==null && root.right==null) {18             if (remain-root.val == 0) return true;19             else return false;20         }21         boolean left=false, right=false;22         if (root.left != null) {23             left = helper(root.left, remain-root.val);24         }25         if (root.right != null) {26             right = helper(root.right, remain-root.val);27         }28         return left || right;29     }30 }

 

 精炼简洁的做法，但是不容易写：

1 public boolean hasPathSum(TreeNode root, int sum) {2     if(root == null)3         return false;4     if(root.left == null && root.right==null && root.val==sum)5         return true;6     return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum-root.val);7 }