Path Sum II - LeetCode

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Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5             / \            4   8           /   / \          11  13  4         /  \    / \        7    2  5   1

return

[   [5,4,11,2],   [5,8,4,5]]

这道题我报的Wrong Answer和我Eclipse自己做的结果居然不一样，我不知道这是Online Judge的什么Bug或者是我的eclipse出了问题。用同样的测试样例，我的eclipse输出的就是正确结果。

代码如下：

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    static ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();static ArrayList<Integer> r = new ArrayList<Integer>();    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {    if(root == null){return result;}else if (sum == root.val && root.left == null && root.right == null){r.add(root.val);result.add(r);r = new ArrayList<Integer>();return result;}else if(root.left == null && root.right == null){return result;}else {if(root.left != null && (sum - root.val) >= 0){r.add(root.val);pathSum(root.left, sum - root.val);if(r.size() >= 1){r.remove(r.size()-1);}}if (root.right !=null && (sum - root.val) >= 0){r.add(root.val);pathSum(root.right, sum - root.val);if(r.size() >= 1){r.remove(r.size()-1);}}}return result;    }}

这里我又借鉴了一些比较简单清晰的代码（经测试AC）：

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public ArrayList<ArrayList<Integer>> pathSum(TreeNode root, int sum) {        // Start typing your Java solution below        // DO NOT write main() function        ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();        if(root == null)            return list;        ArrayList<Integer> tmp = new ArrayList<Integer>();        list = findPath(list, tmp, root, sum);        return list;            }        public ArrayList<ArrayList<Integer>> findPath(ArrayList<ArrayList<Integer>> list, ArrayList<Integer> tmp, TreeNode node, int sum) {                if(node == null) {           return list;        }                tmp.add(node.val);        int nodeSum = sum - node.val;        if((nodeSum == 0) && (node.left == null) && (node.right == null)) {            ArrayList<Integer> tmp1 = new ArrayList<Integer>(tmp);            list.add(tmp1);        }            findPath(list, tmp, node.left, nodeSum);        findPath(list, tmp, node.right, nodeSum);        if(tmp.size() > 0) //这一步比较重要，这让树可以回退地寻找其他路径。            tmp.remove(tmp.size() - 1);        return list;                }}

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