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Fixing Two Swapped nodes of a BST

Tags : tree, bst, geeksforgeeks, cpp, hard

Problem Statement - link #

You are given the root of a binary search tree(BST), where exactly two nodes were swapped by mistake. Fix (or correct) the BST by swapping them back. Do not change the structure of the tree. Note: It is guaranteed that the given input will form BST, except for 2 nodes that will be wrong.

Your Task: You don’t need to take any input. Just complete the function correctBst() that takes root node as parameter. The function should return the root of corrected BST. BST will then be checked by driver code and 0 or 1 will be printed.

Expected Time Complexity: O(n)
Expected Auxiliary Space: O(logn)

Examples #

Example 1:

Input:
       10
     /    \
    5      8
   / \
  2   20
Output: 1
Explanation:

Example 2:

Input:
         11
       /    \
      3      17
       \    /
        4  10
Output: 1
Explanation:
By swapping nodes 11 and 10, the BST
can be fixed.

Constraints #

Solutions #


/*struct Node
{
    int data;
    struct Node *left, *right;
};*/

class Solution {
    void helper(struct Node *root, struct Node *&pre, struct Node *&n1, struct Node *&n2) {
        if(!root) {
            return;
        }
        helper(root->left, pre, n1, n2);
        if(pre && pre->data > root->data) {
            if(!n1) {
                n1 = pre;
            }
            n2 = root;
        }
        pre = root;
        helper(root->right, pre, n1, n2);
    }
  public:
    struct Node *correctBST(struct Node *root) {
        // code here
        Node *pre = NULL, *n1=NULL, *n2=NULL;
        helper(root, pre, n1, n2);
        if(n1 && n2) {
            swap(n1->data, n2->data);
        }
        return root;
    }
};