leetcode: 99. Recover Binary Search Tree

Q

Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

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

AC

# Definition for a binary tree node.class TreeNode(object):    def __init__(self, x):        self.val = x        self.left = None        self.right = Noneclass Solution(object):    def recoverTree(self, root):        """        :type root: TreeNode        :rtype: void Do not return anything, modify root in-place instead.        """        self.lefty = None        self.righty = None        self._minnode = None        self._maxnode = None        self.preorder_visit(root)        self.postorder_visit(root)        # print self.lefty, self.righty        self.lefty.val, self.righty.val = self.righty.val, self.lefty.val    def preorder_visit(self, node):        # first check left and right trees        if node.left:            self.preorder_visit(node.left)            if self.lefty:                return        if self._maxnode and node.val<self._maxnode.val:            self.lefty = self._maxnode            return        if not self._maxnode or node.val > self._maxnode.val:            self._maxnode = node        if node.right:            self.preorder_visit(node.right)    def postorder_visit(self, node):        # first check left and right trees        if node.right:            self.postorder_visit(node.right)            if self.righty:                return        if self._minnode and node.val > self._minnode.val:            self.righty = self._minnode            return        if not self._minnode or node.val < self._minnode.val:            self._minnode = node        if node.left:            self.postorder_visit(node.left)# Time:  O(n)# Space: O(1)class TreeNode:    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 Noneclass Solution2(object):    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] = curif __name__ == "__main__":    root, root.left = TreeNode(0), TreeNode(1)    assert Solution().recoverTree(root) == None

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