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【leetcode刷题笔记】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,

              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：其实很类似这道题：http://www.cnblogs.com/sunshineatnoon/p/3853376.html，也是用递归的方法，在每个root上计算一个sum，表示从树根节点到当前节点得到的和，然后判断当前节点是否是叶节点，如果是，再判断从根节点到当前节点路径上的和是否等于sum，是就找到了要求的路径。

代码如下：

 1 /**
 2  * Definition for binary tree
 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     private boolean hadPath = false;
12     private void hasPathDfs(TreeNode root,int sum,int currSum){
13         if(root == null)
14             return;
15         currSum = currSum + root.val;
16         if(root.left == null && root.right == null && currSum == sum){
17             hadPath = true;
18             return;
19         }
20         hasPathDfs(root.left, sum, currSum);
21         hasPathDfs(root.right, sum, currSum);
22     }
23     public boolean hasPathSum(TreeNode root, int sum) {
24         hasPathDfs(root, sum, 0);
25         return hadPath;
26     }
27 }