897. 递增顺序搜索树

Difficulty
easy
Tags
二叉树
URL
https://leetcode-cn.com/problems/increasing-order-search-tree/
Star
给你一棵二叉搜索树的 root ，请你 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：
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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

法 1 中序遍历

思路
中序遍历，维护一个 head 节点，一个prev节点，不断的让prev.right = cur，同时让cur.left = None
题解
# 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:
        if root is None:
            return None
        
        stack = []
        head = None
        prev = None
        while stack or root:
            if root:
                stack.append(root)
                root = root.left
            else:
                root = stack.pop()
                if head is None:
                    head = root

                if prev:
                    prev.right = root
                    
                prev = root
                prev.left = None
                root = root.right

        return head