Question

426. 将二叉搜索树转化为排序的双向链表

English Version

题目描述

将一个 二叉搜索树 就地转化为一个 已排序的双向循环链表 。

对于双向循环列表，你可以将左右孩子指针作为双向循环链表的前驱和后继指针，第一个节点的前驱是最后一个节点，最后一个节点的后继是第一个节点。

特别地，我们希望可以 就地 完成转换操作。当转化完成以后，树中节点的左指针需要指向前驱，树中节点的右指针需要指向后继。还需要返回链表中最小元素的指针。

 

示例 1：

输入：root = [4,2,5,1,3] 


输出：[1,2,3,4,5]

解释：下图显示了转化后的二叉搜索树，实线表示后继关系，虚线表示前驱关系。

示例 2：

输入：root = [2,1,3]
输出：[1,2,3]

示例 3：

输入：root = []
输出：[]
解释：输入是空树，所以输出也是空链表。

示例 4：

输入：root = [1]
输出：[1]

 

提示：

• -1000 <= Node.val <= 1000
• Node.left.val < Node.val < Node.right.val
• Node.val 的所有值都是独一无二的
• 0 <= Number of Nodes <= 2000

解法

方法一

"""
# Definition for a Node.
class Node:
  def __init__(self, val, left=None, right=None):
    self.val = val
    self.left = left
    self.right = right
"""


class Solution:
  def treeToDoublyList(self, root: 'Optional[Node]') -> 'Optional[Node]':
    def dfs(root):
      if root is None:
        return
      nonlocal prev, head
      dfs(root.left)
      if prev:
        prev.right = root
        root.left = prev
      else:
        head = root
      prev = root
      dfs(root.right)

    if root is None:
      return None
    head = prev = None
    dfs(root)
    prev.right = head
    head.left = prev
    return head

/*
// Definition for a Node.
class Node {
  public int val;
  public Node left;
  public Node right;

  public Node() {}

  public Node(int _val) {
    val = _val;
  }

  public Node(int _val,Node _left,Node _right) {
    val = _val;
    left = _left;
    right = _right;
  }
};
*/

class Solution {
  private Node prev;
  private Node head;

  public Node treeToDoublyList(Node root) {
    if (root == null) {
      return null;
    }
    prev = null;
    head = null;
    dfs(root);
    prev.right = head;
    head.left = prev;
    return head;
  }

  private void dfs(Node root) {
    if (root == null) {
      return;
    }
    dfs(root.left);
    if (prev != null) {
      prev.right = root;
      root.left = prev;
    } else {
      head = root;
    }
    prev = root;
    dfs(root.right);
  }
}

/*
// Definition for a Node.
class Node {
public:
  int val;
  Node* left;
  Node* right;

  Node() {}

  Node(int _val) {
    val = _val;
    left = NULL;
    right = NULL;
  }

  Node(int _val, Node* _left, Node* _right) {
    val = _val;
    left = _left;
    right = _right;
  }
};
*/

class Solution {
public:
  Node* prev;
  Node* head;

  Node* treeToDoublyList(Node* root) {
    if (!root) return nullptr;
    prev = nullptr;
    head = nullptr;
    dfs(root);
    prev->right = head;
    head->left = prev;
    return head;
  }

  void dfs(Node* root) {
    if (!root) return;
    dfs(root->left);
    if (prev) {
      prev->right = root;
      root->left = prev;
    } else
      head = root;
    prev = root;
    dfs(root->right);
  }
};

/**
 * Definition for a Node.
 * type Node struct {
 *   Val int
 *   Left *Node
 *   Right *Node
 * }
 */

func treeToDoublyList(root *Node) *Node {
  if root == nil {
    return root
  }
  var prev, head *Node

  var dfs func(root *Node)
  dfs = func(root *Node) {
    if root == nil {
      return
    }
    dfs(root.Left)
    if prev != nil {
      prev.Right = root
      root.Left = prev
    } else {
      head = root
    }
    prev = root
    dfs(root.Right)
  }
  dfs(root)
  prev.Right = head
  head.Left = prev
  return head
}

/**
 * // Definition for a Node.
 * function Node(val, left, right) {
 *    this.val = val;
 *    this.left = left;
 *    this.right = right;
 *  };
 */

/**
 * @param {Node} root
 * @return {Node}
 */
var treeToDoublyList = function (root) {
  if (!root) return root;
  let prev = null;
  let head = null;

  function dfs(root) {
    if (!root) return;
    dfs(root.left);
    if (prev) {
      prev.right = root;
      root.left = prev;
    } else {
      head = root;
    }
    prev = root;
    dfs(root.right);
  }
  dfs(root);
  prev.right = head;
  head.left = prev;
  return head;
};