【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 }