leetcode 99. Recover Binary Search Tree(恢复二叉搜索树)

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Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Example 1:

Input: [1,3,null,null,2]

   1
  /
 3
  \
   2

Output: [3,1,null,null,2]

   3
  /
 1
  \
   2

二叉搜索树,其中两个节点发生了对换,导致不再是二叉搜索树,要求复原二叉搜索树

二叉搜索树的条件,左子树的值 < root的值 < 右子树的值

思路:

如果用中序遍历二叉搜索树,结果应该是升序的
如果其中出现了违反升序的节点,就说明找到了发生对换的节点

注意题中说只对换了2个element,所以就比较简单一些

比如正常顺序:
1,2,3,4,5,6

而现在中序遍历得到
1,5,3,4,2,6

2本来应该在第2个位置,现在换到第5个,2本来比3小,因为正常应该是升序的,所以也就比3后面的都小,想找到2的位置,只需要找到最后一个前一节点>后一节点的位置
想找到5的位置,只需要找到第一个发生倒序的位置

如果是相邻node交换,那后面应该一路都是升序的,如果后面再次出现降序(倒序),说明前面较小的节点换到后面去了,要换回来,
简而言之,就是要把第一次降序的第一个节点的值,和最后一次降序的第二个节点的值交换。
因为只需要记前后两节点,所以用一个pre记前一个节点,root为当前节点。

*注意java语言是按值传参数的,如果把TreeNode通过参数传递到函数内部,即使函数内部值改变,外部的值也还是不变,所以保存第一个倒序和最后一个倒序的TreeNode节点的定义放在class的global变量

class Solution {
    TreeNode pre;
    TreeNode first;
    TreeNode second;
    
    public void recoverTree(TreeNode root) {
        inOrder(root);
        
        //swap val
        int tmp = first.val;
        first.val = second.val;
        second.val = tmp;
    }
    
    //找到第一次降序的第一个node,和最后一次降序的第二个node
    void inOrder(TreeNode root) {
        if(root == null) return;
        
        inOrder(root.left);
        
        //pre是前一个root,出现降序
        if(pre != null && pre.val > root.val) {
            if(first == null) {  //第一次降序
                first = pre;
            }
            second = root;  //每次降序都记下第二个元素
        }
        pre = root;
        
        inOrder(root.right);
    }
}
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【LeetCode】#99恢复二叉搜索树(Recover Binary Search Tree) 题目描述 二叉搜索树中的两个节点被错误地交换。 请在不改变其结构的情况下,恢复这棵树。 示例 示例 1: 输入: [1,3,null,null,2] 1 / 3 2 输出: [3,1,null,null,2] 3 / 1 2 示例 2: 输入: [3,1,4,null,null,2] 3 / 1...

weixin_43858604的博客 323

[leetcode]99. Recover Binary Search Tree恢复二叉搜索树

Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Example 1: Input: [1,3,null,null,2] 1 / 3 \ 2 Output: [...

weixin_30470857的博客 113

[Swift]LeetCode99. 恢复二叉搜索树 | Recover Binary Search Tree

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★➤微信公众号:山青咏芝(shanqingyongzhi)➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/)➤GitHub地址:https://github.com/strengthen/LeetCode➤原文地址:https://www.cnblogs.com/streng...

weixin_30632089的博客 95

leetcode 99. Recover Binary Search Tree BST二叉搜索树的两元素交换的恢复 + 中序遍历 + 很棒的分析和做法

Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note: A solution using O(n) space is pretty straight forward. Could you devise a

JackZhangNJU的专栏 721

LeetCode 99. Recover Binary Search Tree(修复二叉搜索树)

原题网址:https://leetcode.com/problems/recover-binary-search-tree/ Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure. Note: A sol

James Pan 708

LeetCode-99-Recover Binary Search Tree 二叉树交换结点

题意:给一个有序的二叉树,已知有两个结点被swap了,让你恢复,要求空间复杂度为O1。 题解:中序遍历,应该是递增的,当冲突时记录父子的结点。如果不出意外,会有两次冲突,如果只有一次冲突,那就交换这个父子就好。如果有两次冲突,就交换第一次的父亲和第二次的儿子。 举例: 1234567->7234561,这时交换72组合的7 with 61组合的1 1234567->2134567,没找到第

世靖的码场 459

[Leetcode][python]Recover Binary Search Tree/恢复二叉搜索树

题目大意一颗二叉查找树中的某两个节点被错误的交换了,需要恢复成原来的正确的二叉查找树。挑战:是要求空间复杂度为常数空间。空间复杂度为O(N)是常规解法。解题思路来自:博客算法一: 思路很简单,一颗二叉查找树的中序遍历应该是升序的,而两个节点被交换了,那么对这个错误的二叉查找树中序遍历,肯定不是升序的。那我们只需把顺序恢复过来然后进行重新赋值就可以了。(可针对任意个数目的节点错乱的情况)算法二:

蛮三刀酱 1428

[leetcode]99. Recover Binary Search Tree@Java解题报告

https://leetcode.com/problems/recover-binary-search-tree/description/ Two elements of a binary search tree (BST) are swapped by mistake. Recover the tree without changing its structure.

zjkC050818的博客 608
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