Binary Search Tree Typo - Microsoft Top Interview Questions


Problem Statement :


You are given a binary tree root which is almost a binary search tree except two nodes' values have been swapped. Return the original binary search tree.

Constraints

n ≤ 100,000 where n is the number of nodes in root

Example 1

Input

root = [2, [5, null, null], [7, [1, null, null], [8, null, null]]]

Output

[2, [1, null, null], [7, [5, null, null], [8, null, null]]]

Explanation

We can swap 1 and 5.




Example 2

Input




root = [0, [1, null, null], null]

Output



[1, [0, null, null], null]

Explanation

We can swap 0 and 1.



Solution :



title-img




                        Solution in C++ :

void check(Tree* root, Tree*& prev, Tree*& node1, Tree*& node2) {
    if (!root) return;
    check(root->left, prev, node1, node2);
    if (prev && prev->val >= root->val) {
        if (node1 == nullptr) node1 = prev;
        node2 = root;
    }
    prev = root;
    check(root->right, prev, node1, node2);
}

Tree* solve(Tree* root) {
    Tree *node1 = nullptr, *node2 = nullptr, *prev = nullptr;
    check(root, prev, node1, node2);
    swap(node1->val, node2->val);
    return root;
}
                    


                        Solution in Java :

import java.util.*;

/**
 * public class Tree {
 *   int val;
 *   Tree left;
 *   Tree right;
 * }
 */
class Solution {
    public Tree solve(Tree root) {
        List<Tree> values = new ArrayList<>();
        inOrder(root, values);
        findAndSwap(values);
        return root;
    }

    private void inOrder(Tree node, List<Tree> values) {
        if (node == null)
            return;

        inOrder(node.left, values);
        values.add(node);
        inOrder(node.right, values);
    }

    private void findAndSwap(List<Tree> values) {
        Tree first = null;
        Tree second = null;

        for (int i = 1; i < values.size(); i++) {
            if (values.get(i - 1).val > values.get(i).val) {
                if (first == null) {
                    first = values.get(i - 1);
                    second = values.get(i);
                } else {
                    second = values.get(i);
                }
            }
        }

        int temp = first.val;
        first.val = second.val;
        second.val = temp;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, root):
        left = right = None
        last_node = Tree(-float("inf"))

        for node in self.morris_inorder(root):
            if node.val < last_node.val:
                left = left or last_node
                right = node
            last_node = node

        left.val, right.val = right.val, left.val

        return root

    @staticmethod
    def morris_inorder(node):
        temp = None
        while node:
            if node.left:
                temp = node.left
                while temp.right and temp.right != node:
                    temp = temp.right
                if temp.right:
                    temp.right = None
                    yield node
                    node = node.right
                else:
                    temp.right = node
                    node = node.left
            else:
                yield node
                node = node.right
                    


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