99. Recover Binary Search Tree

problem description

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

Follow up:

• A solution using O(n) space is pretty straight forward.

• Could you devise a constant space solution?

algorithm thought

已知BST中有两个数字是错的，那么我们就找到两个错的位置就行。还是中序遍历，找到不是顺序增长的两个位置。记录下来，最后交换两个的值

code

/**
 * 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* first=NULL;
    TreeNode* second=NULL;
    int count=0;
    TreeNode* pre=new TreeNode(INT_MIN);
    void recoverTree(TreeNode* root) {
        bool bo=false;
        travers(root);
        int tmp=first->val;
        first->val=second->val;
        second->val=tmp;
    }
    void travers(TreeNode* root){
        if(count==2||root==NULL)
            return;
        travers(root->left);
        
        if(first==NULL&&pre->val>root->val){
            first=pre;
        }
        
        if(first!=NULL&&pre->val>root->val){
            second=root;
            count++;
        }
        pre=root;
        travers(root->right);
    }
};

//利用bst的中序遍历性质来解决问题

algorithm analysis

一次遍历，时间复杂度O(n)