0099._Recover_Binary_Search_Tree.md

99. Recover Binary Search Tree

难度:Hard

刷题内容

原题连接

• https://leetcode.com/problems/recover-binary-search-tree/

内容描述

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
Example 2:

Input: [3,1,4,null,null,2]

  3
 / \
1   4
   /
  2

Output: [2,1,4,null,null,3]

  2
 / \
1   4
   /
  3

思路 - 时间复杂度: O(n)- 空间复杂度: O(1)******

这道首先要搞懂平衡二叉树在中序遍历时递增的。清楚这个性质后我们只需要对二叉树进行中序遍历，找到哪个错误的数即可，比如第一个例子中序遍历之后为 321，交换3和1，第二个例子中序遍历之后为 1324，交换3和2。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
TreeNode* n1;
TreeNode* n2;
TreeNode* pre = new TreeNode(INT_MIN);
void travel(TreeNode* root){
      if(!root)
        return;
    travel(root ->left);
    if(!n1 && root ->val <= pre ->val)
        n1 = pre;
    if(pre && n1 && root ->val <= pre ->val)
        n2 = root;
    pre = root;
    travel(root ->right);
}
void recoverTree(TreeNode* root) {
    travel(root);
    swap(n1 ->val,n2 ->val);
}
};