本问题对应的 leetcode 原文链接：剑指 Offer 36. 二叉搜索树与双向链表
问题描述
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。
解题思路
视频讲解直达： 本题视频讲解
代码实现
class Solution {
public Node treeToDoublyList(Node root) {
if(root == null){
return null;
}
Queue<Node> queue = new LinkedList<>();
inOrder(root, queue);
Node head = queue.poll();
Node pre = head;
while(!queue.isEmpty()){
Node cur = queue.poll();
pre.right = cur;
cur.left = pre;
pre = cur;
}
pre.right = head;
head.left = pre;
return head;
}
void inOrder(Node root, Queue<Node> queue){
if(root == null){
return;
}
inOrder(root.left, queue);
queue.add(root);
inOrder(root.right, queue);
}
}
Python
# Definition for a Node.
class Node(object):
def __init__(self, val, left=None, right=None):
self.val = val
self.left = left
self.right = right
class Solution(object):
def treeToDoublyList(self, root):
"""
:type root: Node
:rtype: Node
"""
if not root:
return None
queue = []
self.inOrder(root, queue)
head = queue.pop(0)
pre = head
while queue:
cur = queue.pop(0)
pre.right = cur
cur.left = pre
pre = cur
pre.right = head
head.left = pre
return head
def inOrder(self, root, queue):
if not root:
return
self.inOrder(root.left, queue)
queue.append(root)
self.inOrder(root.right, queue)
C++
/*
// 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;
}
vector<Node*> queue;
inOrder(root, queue);
Node* head = queue[0];
Node* pre = head;
for (int i = 1; i < queue.size(); i++) {
Node* cur = queue[i];
pre->right = cur;
cur->left = pre;
pre = cur;
}
pre->right = head;
head->left = pre;
return head;
}
void inOrder(Node* root, vector<Node*>& queue) {
if (!root) {
return;
}
inOrder(root->left, queue);
queue.push_back(root);
inOrder(root->right, queue);
}
};
Go
/**
* Definition for a Node.
* type Node struct {
* Val int
* Next *Node
* Random *Node
* }
*/
func copyRandomList(head *Node) *Node {
if head == nil {
return nil
}
// 复制链表节点
cur := head
for cur != nil {
next := cur.Next
cur.Next = &Node{Val: cur.Val}
cur.Next.Next = next
cur = next
}
// 复制随机节点
cur = head
for cur != nil {
curNew := cur.Next
if cur.Random != nil {
curNew.Random = cur.Random.Next
}
cur = cur.Next.Next
}
// 拆分,比如把 A->A1->B->B1->C->C1拆分成 A->B->C和A1->B1->C1
headNew := head.Next
cur = head
curNew := head.Next
for cur != nil {
cur.Next = cur.Next.Next
cur = cur.Next
if cur == nil {
curNew.Next = nil
} else {
curNew.Next = cur.Next
}
curNew = curNew.Next
}
return headNew
}
JS
/**
* // 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 == null) {
return null;
}
var queue = [];
inOrder(root, queue);
var head = queue.shift();
var pre = head;
while (queue.length !== 0) {
var cur = queue.shift();
pre.right = cur;
cur.left = pre;
pre = cur;
}
pre.right = head;
head.left = pre;
return head;
function inOrder(root, queue) {
if (root == null) {
return;
}
inOrder(root.left, queue);
queue.push(root);
inOrder(root.right, queue);
}
};
shuaidi