99. Recover Binary Search Tree

Description

You are given the root of a binary search tree (BST), where the values of exactly two nodes of the tree were swapped by mistake. Recover the tree without changing its structure.

 

Example 1:

Input: root = [1,3,null,null,2]
Output: [3,1,null,null,2]
Explanation: 3 cannot be a left child of 1 because 3 > 1. Swapping 1 and 3 makes the BST valid.

Example 2:

Input: root = [3,1,4,null,null,2]
Output: [2,1,4,null,null,3]
Explanation: 2 cannot be in the right subtree of 3 because 2 < 3. Swapping 2 and 3 makes the BST valid.

 

Constraints:

• The number of nodes in the tree is in the range [2, 1000].
• -231 <= Node.val <= 231 - 1

 

Follow up: A solution using O(n) space is pretty straight-forward. Could you devise a constant O(1) space solution?

Solution

recover-binary-search-tree.py

class Solution(object): 
    def recoverTree(self, root): # O(lg(n)) space
        pre = first = second = None
        stack = []
        while True:
            while root:
                stack.append(root)
                root = root.left
            if not stack:
                break
            node = stack.pop()
            if not first and pre and pre.val > node.val:
                first = pre
            if first and pre and pre.val > node.val:
                second = node
            pre = node
            root = node.right
        first.val, second.val = second.val, first.val

    def recoverTree1(self, root): # O(n+lg(n)) space  
        res = []
        self.dfs(root, res)
        first, second = None, None
        for i in range(len(res)-1):
            if res[i].val > res[i+1].val and not first:
                first = res[i]
            if res[i].val > res[i+1].val and first:
                second = res[i+1]
        first.val, second.val = second.val, first.val

    def dfs(self, root, res):
        if root:
            self.dfs(root.left, res)
            res.append(root)
            self.dfs(root.right, res)