112. Path Sum

问题描述：
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,

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

自己写的有误的代码,正确的代码在后面：

/** * 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 && sum==0)            return true;        if(root==null && sum!=0)            return false;        return hasPathSum(root.left,sum-root.val)||hasPathSum(root.right,sum-root.val);    }}

正确的java代码:

/** * 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 && root.val == sum) return true;        return hasPathSum(root.left,sum-root.val)||hasPathSum(root.right,sum-root.val);    }}

总结：对于递归的边界要特别小心处理，否则可能会转晕。