Problem description:

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

Recover the tree without changing its structure.

Example 1:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Input: [1,3,null,null,2]

   1
  /
 3
  \
   2

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

   3
  /
 1
  \
   2

Example 2:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
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?

Solution:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
/**
 * 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, *second; //store the position that needs to be swap
    TreeNode *prev; // store the previous position in tree
    void recoverTree(TreeNode* root) {
        first = second = prev = NULL;
        //inorder traversal
        dfs(root);
        if (first != NULL && second != NULL) {
            int tmp = first -> val;
            first -> val = second -> val;
            second -> val = tmp;
        }
    }
    void dfs(TreeNode* root) {
        // check left first
        if (root -> left != NULL) dfs(root -> left);
        
        // if there's any disorder, store the position. 
        if (prev != NULL && prev -> val > root -> val) {
            if (first == NULL) 
                first = prev;
            if (first != NULL) 
                second = root;
        }
        // current root become prev
        prev = root;
        // right subtree
        if (root -> right != NULL) dfs(root -> right);
    }
};

time complexity: $O(n)$
space complexity: $O(logn)$

Solution2

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
/**
 * 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, *second= NULL;
    TreeNode *prev= new TreeNode(INT_MIN);
    void recoverTree(TreeNode* root) {
        inorder(root);
        swap(first->val, second->val);
    }
    
    void inorder(TreeNode* root){
        if(!root) return;
        inorder(root->left);
        
        //do something
        if(first == NULL && prev->val >= root->val) first= prev;
        if(first != NULL && prev->val >= root->val) second= root;
        //end do something
        prev= root;
        inorder(root->right);
    }
};

reference:
https://goo.gl/35YynW
https://goo.gl/SVX1A6