Question

面试题 36. 二叉搜索树与双向链表

题目描述

输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的循环双向链表。要求不能创建任何新的节点，只能调整树中节点指针的指向。

 

为了让您更好地理解问题，以下面的二叉搜索树为例：

 

 

我们希望将这个二叉搜索树转化为双向循环链表。链表中的每个节点都有一个前驱和后继指针。对于双向循环链表，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。

下图展示了上面的二叉搜索树转化成的链表。“head” 表示指向链表中有最小元素的节点。

 

 

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。还需要返回链表中的第一个节点的指针。

 

注意：本题与主站 426 题相同：https://leetcode.cn/problems/convert-binary-search-tree-to-sorted-doubly-linked-list/

注意：此题对比原题有改动。

解法

方法一：中序遍历

二叉搜索树的中序遍历是有序序列，因此可以通过中序遍历得到有序序列，过程中构建双向链表。

遍历结束，将头节点和尾节点相连，返回头节点。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉搜索树的节点个数。

"""
# 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: "Node") -> "Node":
    def dfs(root):
      if root is None:
        return
      dfs(root.left)
      nonlocal head, pre
      if pre:
        pre.right = root
      else:
        head = root
      root.left = pre
      pre = root
      dfs(root.right)

    if root is None:
      return None
    head = pre = None
    dfs(root)
    head.left = pre
    pre.right = head
    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 head;
  private Node pre;

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

  private void dfs(Node root) {
    if (root == null) {
      return;
    }
    dfs(root.left);
    if (pre != null) {
      pre.right = root;
    } else {
      head = root;
    }
    root.left = pre;
    pre = 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* treeToDoublyList(Node* root) {
    if (!root) {
      return nullptr;
    }
    Node* pre = nullptr;
    Node* head = nullptr;
    function<void(Node*)> dfs = [&](Node* root) {
      if (!root) {
        return;
      }
      dfs(root->left);
      if (pre) {
        pre->right = root;
      } else {
        head = root;
      }
      root->left = pre;
      pre = root;
      dfs(root->right);
    };

    dfs(root);
    head->left = pre;
    pre->right = head;
    return head;
  }
};

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

func treeToDoublyList(root *Node) *Node {
  if root == nil {
    return nil
  }
  var head, pre *Node
  var dfs func(*Node)
  dfs = func(root *Node) {
    if root == nil {
      return
    }
    dfs(root.Left)
    if pre != nil {
      pre.Right = root
    } else {
      head = root
    }
    root.Left = pre
    pre = root
    dfs(root.Right)
  }
  dfs(root)
  head.Left = pre
  pre.Right = head
  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 null;
  }
  let head = null;
  let pre = null;
  const dfs = root => {
    if (!root) {
      return;
    }
    dfs(root.left);
    if (pre) {
      pre.right = root;
    } else {
      head = root;
    }
    root.left = pre;
    pre = root;
    dfs(root.right);
  };
  dfs(root);
  head.left = pre;
  pre.right = head;
  return head;
};

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

  public Node() {}

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

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

public class Solution {
  private Node head;
  private Node pre;

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

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