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给你一棵二叉搜索树，请 按中序遍历 将其重新排列为一棵递增顺序搜索树，使树中最左边的节点成为树的根节点，并且每个节点没有左子节点，只有一个右子节点。
示例 1：
输入：root = [5,3,6,2,4,null,8,1,null,null,null,7,9] 输出：[1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]
示例 2：
输入：root = [5,1,7] 输出：[1,null,5,null,7]
提示：
- 树中节点数的取值范围是
[1, 100] 0 <= Node.val <= 1000
注意：本题与主站 897 题相同： https://leetcode.cn/problems/increasing-order-search-tree/
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def increasingBST(self, root: TreeNode) -> TreeNode:
head, tail = None, None
stack = []
cur = root
while stack or cur:
while cur:
stack.append(cur)
cur = cur.left
cur = stack.pop()
if not head:
head = cur
else:
tail.right = cur
tail = cur
cur.left = None
cur = cur.right
return head/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode increasingBST(TreeNode root) {
TreeNode head = null, tail = null;
Deque<TreeNode> stack = new ArrayDeque<>();
TreeNode cur = root;
while (!stack.isEmpty() || cur != null) {
while (cur != null) {
stack.push(cur);
cur = cur.left;
}
cur = stack.pop();
if (head == null) {
head = cur;
} else {
tail.right = cur;
}
tail = cur;
cur.left = null;
cur = cur.right;
}
return head;
}
}/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* increasingBST(TreeNode* root) {
TreeNode *head = nullptr, *tail = nullptr;
stack<TreeNode*> stk;
TreeNode* cur = root;
while (!stk.empty() || cur != nullptr) {
while (cur != nullptr) {
stk.push(cur);
cur = cur->left;
}
cur = stk.top();
stk.pop();
if (head == nullptr) {
head = cur;
} else {
tail->right = cur;
}
tail = cur;
cur->left = nullptr;
cur = cur->right;
}
return head;
}
};/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func increasingBST(root *TreeNode) *TreeNode {
var head, tail *TreeNode
stack := make([]*TreeNode, 0)
cur := root
for len(stack) > 0 || cur != nil {
for cur != nil {
stack = append(stack, cur)
cur = cur.Left
}
cur = stack[len(stack)-1]
stack = stack[:len(stack)-1]
if head == nil {
head = cur
} else {
tail.Right = cur
}
tail = cur
cur.Left = nil
cur = cur.Right
}
return head
}/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function increasingBST(root: TreeNode | null): TreeNode | null {
const dummy = new TreeNode();
let cur = dummy;
const dfs = (root: TreeNode | null) => {
if (root == null) {
return;
}
dfs(root.left);
cur.right = new TreeNode(root.val);
cur = cur.right;
dfs(root.right);
};
dfs(root);
return dummy.right;
}// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, vals: &mut Vec<i32>) {
if root.is_none() {
return;
}
let node = root.as_ref().unwrap().borrow();
Self::dfs(&node.left, vals);
vals.push(node.val);
Self::dfs(&node.right, vals);
}
pub fn increasing_bst(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {
let mut vals = Vec::new();
Self::dfs(&root, &mut vals);
let mut dummy = Rc::new(RefCell::new(TreeNode::new(0)));
for &val in vals.iter().rev() {
let mut dummy = dummy.as_ref().borrow_mut();
dummy.right = Some(Rc::new(RefCell::new(TreeNode {
val,
left: None,
right: dummy.right.take(),
})));
}
let ans = dummy.as_ref().borrow_mut().right.take();
ans
}
}/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
struct TreeNode* dfs(struct TreeNode* root, struct TreeNode* cur) {
if (!root) {
return cur;
}
cur = dfs(root->left, cur);
cur->right = malloc(sizeof(struct TreeNode));
cur->right->val = root->val;
cur->right->left = NULL;
cur->right->right = NULL;
cur = cur->right;
return dfs(root->right, cur);
}
struct TreeNode* increasingBST(struct TreeNode* root) {
struct TreeNode* dummy = malloc(sizeof(struct TreeNode));
dfs(root, dummy);
return dummy->right;
}/* class TreeNode {
* var val: Int
* var left: TreeNode?
* var right: TreeNode?
* init() {
* self.val = 0
* self.left = nil
* self.right = nil
* }
* init(_ val: Int) {
* self.val = val
* self.left = nil
* self.right = nil
* }
* init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
* self.val = val
* self.left = left
* self.right = right
* }
* }
*/
class Solution {
func increasingBST(_ root: TreeNode?) -> TreeNode? {
var head: TreeNode? = nil
var tail: TreeNode? = nil
var stack = [TreeNode]()
var cur = root
while !stack.isEmpty || cur != nil {
while cur != nil {
stack.append(cur!)
cur = cur?.left
}
cur = stack.removeLast()
if head == nil {
head = cur
} else {
tail?.right = cur
}
tail = cur
cur?.left = nil
cur = cur?.right
}
return head
}
}# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def increasingBST(self, root: TreeNode) -> TreeNode:
def dfs(root: TreeNode):
if root is None:
return
dfs(root.left)
nonlocal cur
cur.right = root
root.left = None
cur = cur.right
dfs(root.right)
cur = dummy = TreeNode()
dfs(root)
return dummy.right/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private TreeNode cur;
public TreeNode increasingBST(TreeNode root) {
TreeNode dummy = new TreeNode();
cur = dummy;
dfs(root);
return dummy.right;
}
private void dfs(TreeNode root) {
if (root == null) {
return;
}
dfs(root.left);
cur.right = root;
root.left = null;
cur = cur.right;
dfs(root.right);
}
}/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* increasingBST(TreeNode* root) {
TreeNode* dummy = new TreeNode();
TreeNode* cur = dummy;
function<void(TreeNode*)> dfs = [&](TreeNode* root) {
if (!root) {
return;
}
dfs(root->left);
cur->right = root;
root->left = nullptr;
cur = cur->right;
dfs(root->right);
};
dfs(root);
return dummy->right;
}
};/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func increasingBST(root *TreeNode) *TreeNode {
dummy := &TreeNode{}
cur := dummy
var dfs func(*TreeNode)
dfs = func(root *TreeNode) {
if root == nil {
return
}
dfs(root.Left)
root.Left = nil
cur.Right = root
cur = root
dfs(root.Right)
}
dfs(root)
return dummy.Right
}