| comments | difficulty | edit_url |
|---|---|---|
true |
中等 |
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。
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
"""
# 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);
}
}/* Definition for a Node.
* public class Node {
* public var val: Int
* public var left: Node?
* public var right: Node?
* public init() {
* self.val = 0
* self.left = nil
* self.right = nil
* }
* public init(_ val: Int) {
* self.val = val
* self.left = nil
* self.right = nil
* }
* public init(_ val: Int, _ left: Node?, _ right: Node?) {
* self.val = val
* self.left = left
* self.right = right
* }
* }
*/
class Solution {
private var head: Node?
private var pre: Node?
func treeToDoublyList(_ root: Node?) -> Node? {
if root == nil {
return nil
}
dfs(root)
head?.left = pre
pre?.right = head
return head
}
private func dfs(_ root: Node?) {
guard let root = root else {
return
}
dfs(root.left)
if let preNode = pre {
preNode.right = root
} else {
head = root
}
root.left = pre
pre = root
dfs(root.right)
}
}