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]
]


方法1：只要在每个树节点中加入一个父节点，就可以用上一题的解法了。
方法2：dfs

方法1的代码：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
 
    class TreeNodePlus {
        TreeNode node;
        TreeNodePlus father;
        TreeNodePlus(TreeNode x,TreeNodePlus f) {
            node=x;
            father=f;
        }
    }

    List<List<Integer>> paths = new ArrayList<List<Integer>>();

    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        TreeNodePlus newRoot = new TreeNodePlus(root, null);
        hasPathSum(newRoot, sum);
        return paths;
    }

    public void hasPathSum(TreeNodePlus root, int sum){
        if (root.node == null) {
            return ;
        }
        if (root.node.val==sum && root.node.left==null && root.node.right==null) {
            ArrayList<Integer> list = new ArrayList<Integer>();
            TreeNodePlus father = root;
            while (father != null) {
                list.add(0,father.node.val);
                father = father.father;

            }
            paths.add(list);
            return;
        }

        hasPathSum(new TreeNodePlus(root.node.left,root), sum-root.node.val) ;
        hasPathSum(new TreeNodePlus(root.node.right,root), sum-root.node.val);
    }

}

　　方法2的代码

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
 


    List<List<Integer>> paths = new ArrayList<List<Integer>>();
    public List<List<Integer>> pathSum(TreeNode root, int sum) {

        List<Integer> list = new ArrayList<Integer>();
        dfs(root, sum, 0, list);
        return paths;
    }
    void dfs(TreeNode root, int sum, int curr, List<Integer> list) {
        if (root == null) {
            return;
        }

        if (root.left == null && root.right == null) {
            if (curr + root.val == sum) {
                list.add(root.val);
                paths.add(list);

            }
            return;
        }

        list.add(root.val);
        dfs(root.left, sum, curr + root.val, new ArrayList<Integer>(list));

        dfs(root.right, sum, curr + root.val, new ArrayList<Integer>(list));
    }

}