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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 }
【leetcode刷题笔记】Path Sum,布布扣,bubuko.com
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原文地址：http://www.cnblogs.com/sunshineatnoon/p/3853407.html
