算法系列——Path Sum

题目描述

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.

解题思路

基本的深度优先搜索，记录下当前已经累加到的值，递归调用，到达叶子节点时判断 和是否与指定的值相同。

程序实现

 * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public boolean hasPathSum(TreeNode root, int sum) {        return dfs(root,sum,0);    }    boolean dfs(TreeNode node,int sum,int curSum){        if(node==null)            return false;        if(node.left==null&&node.right==null)            return curSum+node.val==sum;        return dfs(node.left,sum,curSum+node.val)||dfs(node.right,sum,curSum+node.val);    }}