Leetcode---Recover Binary Search Tree

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本文介绍了一种方法来恢复二叉搜索树，其中两个元素因误操作而交换位置，通过中序遍历来定位并修复这些元素，而无需改变树的结构。提供了一个O(n)空间解决方案的变体，实现常数空间复杂度。

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 constant space solution?


confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.


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Have you met this question in a real interview?

二叉排序树，中序遍历的时候应该输出一个有序序列。但是当反转的时候，就会出现当前节点小于前一节点的情况。记录第一个反转的节点和最后一个反转的节点，然后简单的互换节点内容即可，而无需互换节点本身。


  1. TreeNode * pre=NULL;
  2. TreeNode * p1=NULL;
  3. TreeNode * p2=NULL;
  4. void recoverTreeHelper(TreeNode *root) {
  5.     if(root==NULL)
  6.         return;
  7.     recoverTreeHelper(root->left);
  8.     if(pre==NULL)
  9.         pre=root;
  10.     else{
  11.         if(root->val<pre->val){
  12.             if(p1==NULL)
  13.                 p1=pre;
  14.             p2=root;
  15.         }
  16.         pre=root;
  17.     }
  18.     recoverTreeHelper(root->right);
  19. }
  20. void recoverTree(TreeNode *root) {
  21.     recoverTreeHelper(root);
  22.     if(p1!=NULL && p2!=NULL)
  23.         swap(p1->val,p2->val);
  24. }





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