[LeetCode]114. Flatten Binary Tree to Linked List

https://leetcode.com/problems/flatten-binary-tree-to-linked-list/

将二叉树变成一条线每个节点只有一个右子节点，线的顺序为中序遍历

我的解：

函数返回已经flatten的数组，里面两个元素，是线的开头和结尾

public class Solution {    public void flatten(TreeNode root) {        if (root == null) {            return;        }        dfs(root);    }    private TreeNode[] dfs(TreeNode root) {        TreeNode[] res = {root, root};        TreeNode[] left;        TreeNode[] right;        TreeNode leftNode = root.left;        TreeNode rightNode = root.right;        root.left = null;        if (leftNode != null) {         left = dfs(leftNode);         res[1].right = left[0];         res[1] = left[1];        }        if (rightNode != null) {         right = dfs(rightNode);         res[1].right = right[0];         res[1] = right[1];        }        return res;    }}

Discuss最优解：

用参数pre记录当前要flatten的直线右下方所要连接的节点，flatten返回当前flatten后的直线左上角的头结点

public class Solution {    public void flatten(TreeNode root) {        flatten(root, null);    }    private TreeNode flatten(TreeNode root, TreeNode pre) {        if (root == null) {            return pre;        }        // 先右后左                pre = flatten(root.right, pre);        pre = flatten(root.left, pre);        root.right = pre;        root.left = null;        pre = root;        return pre;    }}