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LC-0099 Medium LeetCode

99. Recover Binary Search Tree

Read the full problem statement on LeetCode.
Difficulty: medium Acceptance: 56% Topics: Tree, Depth-First Search, Binary Search Tree, Binary Tree
View full problem on LeetCode
Reference solution (spoiler · python)
# Time:  O(n)
# Space: O(1)

class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

    def __repr__(self):
        if self:
            serial = []
            queue = [self]

            while queue:
                cur = queue[0]

                if cur:
                    serial.append(cur.val)
                    queue.append(cur.left)
                    queue.append(cur.right)
                else:
                    serial.append("#")

                queue = queue[1:]

            while serial[-1] == "#":
                serial.pop()

            return repr(serial)

        else:
            return None

class Solution(object):
    # @param root, a tree node
    # @return a tree node
    def recoverTree(self, root):
        return self.MorrisTraversal(root)

    def MorrisTraversal(self, root):
        if root is None:
            return
        broken = [None, None]
        pre, cur = None, root

        while cur:
            if cur.left is None:
                self.detectBroken(broken, pre, cur)
                pre = cur
                cur = cur.right
            else:
                node = cur.left
                while node.right and node.right != cur:
                    node = node.right

                if node.right is None:
                    node.right =cur
                    cur = cur.left
                else:
                    self.detectBroken(broken, pre, cur)
                    node.right = None
                    pre = cur
                    cur = cur.right

        broken[0].val, broken[1].val = broken[1].val, broken[0].val

        return root

    def detectBroken(self, broken, pre, cur):
        if pre and pre.val > cur.val:
            if broken[0] is None:
                broken[0] = pre
            broken[1] = cur

Solution from kamyu104/LeetCode-Solutions · MIT