[leetcode] Path Sum II

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]
]

https://oj.leetcode.com/problems/path-sum-ii/

思路：要输出所有结果，dfs枚举。

class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> tmp = new ArrayList<>();
        
        pathSumHelper(root,tmp,result,sum);
        
        return result;
        
    }
    
    private void pathSumHelper(TreeNode root, List<Integer> tmp, List<List<Integer>> result, int sum){
        if(root == null)
            return;
        if(root.left==null && root.right==null && root.val == sum){
            tmp.add(root.val);
            result.add(new ArrayList<Integer>(tmp));
            tmp.remove(tmp.size()-1);
            return;
        }
        
        tmp.add(root.val);
        pathSumHelper(root.left,tmp,result,sum-root.val);
        pathSumHelper(root.right,tmp,result,sum-root.val);
        tmp.remove(tmp.size()-1);
    }
    
}

 第三遍记录：最好采用第二遍的方法，在叶子节点的时候就判断是否满足条件，但是注意此时该节点并未加入到tmp中去，所以需要tmp中先加入root.val，然后 res中添加tmp，最后tmo中的root.val要删掉。

如果在叶子节点的子节点，null节点判断（此时判断n==0即可），但是左右null节点都会满足条件而产生两边结果。

以下方法会有重复元素

import java.util.ArrayList;
import java.util.List;

public class Solution {

    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        if (root == null)
            return res;
        List<Integer> tmp = new ArrayList<Integer>();
        pathSumHelper(root, sum, res, tmp);
        return res;
    }

    private void pathSumHelper(TreeNode root, int sum, List<List<Integer>> res, List<Integer> tmp) {
        if (root == null) {
            if (sum == 0)
                res.add(new ArrayList<Integer>(tmp));
            return;
        }

        tmp.add(root.val);
        pathSumHelper(root.left, sum - root.val, res, tmp);
        pathSumHelper(root.right, sum - root.val, res, tmp);
        tmp.remove(tmp.size() - 1);

    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(5);
        root.left = new TreeNode(4);
        root.left.left = new TreeNode(11);
        root.right = new TreeNode(8);
        root.right.right = new TreeNode(7);

        System.out.println(new Solution().pathSum(root, 20));
    }
}