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。

Python3JavaC++GoJavaScriptC#

"""
# 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);
    }
}

面试题 36. 二叉搜索树与双向链表

题目描述

解法

方法一：中序遍历

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