#112 Path Sum

题目:

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：

这是我开始的解法：

/** * 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) {        if(root==null)            return false;        if(root.left==null && root.right==null)            return sum==root.val;        return hasPathSum(root.left,sum-root.val) || hasPashSum(root.right,sum-root.val);    }}

Submission Result: Compile ErrorMore Details Line 16: error: cannot find symbol: method hasPashSum(TreeNode,int)

好2，肯定要写在一个函数里的啊，怎么想的。

正解：

/** * 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 dfsPath(root,sum);    }        public boolean dfsPath(TreeNode root, int sum){        if(root==null)            return false;        if(root.left==null && root.right==null)            return sum==root.val;        return dfsPath(root.left,sum-root.val) || dfsPath(root.right,sum-root.val);    }}