Path Sum

递归解决方案/** * 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;        else if(root.left == null && root.right == null && sum - root.val == 0)             return true;        else            return (hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val));    }}