【LeetCode OJ】Recover Binary Search Tree

Problem Link:

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

We know that the inorder traversal of a binary search tree should be a sorted array. Therefore, we can compare each node with its previous node in the inorder to find the two swapped elements.

There are at most two cases that current_node < prev_node:

If the two swapped elements are adjacent, then there is only one occurrence that current_node < prev_node. The two elements are just the current_node and prev_node.
If the two swapped elements are not neighbors, then there are two occurrences that current_node < prev_node. The two elements are the prev_node in the first occurrence and the current_node in the second occurrence.

The code is as follows.

# Definition for a  binary tree node

# class TreeNode:

#     def __init__(self, x):

#         self.val = x

#         self.left = None

#         self.right = None

class Solution:

    # @param root, a tree node

    # @return a tree node

    def recoverTree(self, root):

        """

        Traverse the tree in the order of inorder and compare the node to its previous node.

        There are at most two occurrences that prev_node < current_node, if there are two elements swapped.

        If there is only one occurrence, then the two swapped nodes are prev_node and current_node.

        If there are two occurrences, then the two swapped nodes are prev_node in the first occurrence and current_node in the second occurrence.

        """

        prev_node = None

        current_node = None

        stack = []

        p = root

        res1 = None

        res2 = None

        while stack or p:

            if p:

                stack.append(p)

                p = p.left

            else:

                current_node = stack.pop()

                # Compare current node and previous node

                if prev_node:

                    if prev_node.val > current_node.val:

                        if res1:

                            res2 = current_node

                            break

                        else:

                            res1 = prev_node

                            res2 = current_node

                prev_node = current_node

                p = current_node.right

        res1.val, res2.val = res2.val, res1.val

        return root

巴特西

【LeetCode OJ】Recover Binary Search Tree

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