leetcode: 98. Validate Binary Search Tree

Q

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

The left subtree of a node contains only nodes with keys less than the node’s key.
The right subtree of a node contains only nodes with keys greater than the node’s key.
Both the left and right subtrees must also be binary search trees.

Example 1:

    2   / \  1   3

Binary tree [2,1,3], return true.

Example 2:

    1   / \  2   3

Binary tree [1,2,3], return false.

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 isValidBST(self, root):        """        :type root: TreeNode        :rtype: bool        """        if not root:            return True        a, b, c= self.recurse(root)        return c    def recurse(self, root):        if not root.left and not root.right:            return (root.val, root.val, True)        min_ = root.val        max_= root.val        if root.left:            if root.val<=root.left.val:                return (None, None, False)            else:                _min, _max, ok = self.recurse(root.left)                if not ok:                    return (None, None, False)                if _max>=root.val:                    return (None, None, False)                min_= min(min_, _min)        if root.right:            if root.val>=root.right.val:                return (None, None, False)            else:                _min, _max, ok = self.recurse(root.right)                if not ok:                    return (None, None, False)                if _min<=root.val:                    return (None, None, False)                max_= max(max_, _max)        return (min_, max_, True)# Time:  O(n)# Space: O(1)# Definition for a  binary tree nodeclass TreeNode:    def __init__(self, x):        self.val = x        self.left = None        self.right = None# Morris Traversal Solutionclass Solution2(object):    def isValidBST(self, root):        prev, cur = None, root        while cur:            if cur.left is None:                if prev and prev.val >= cur.val:                    return False                prev = 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:                    if prev and prev.val >= cur.val:                        return False                    node.right = None                    prev = cur                    cur = cur.right        return True# Time:  O(n)# Space: O(h)class Solution3(object):    def isValidBST(self, root):        return self.isValidBSTRecu(root, float("-inf"), float("inf"))    def isValidBSTRecu(self, root, low, high):        if root is None:            return True        return low < root.val and root.val < high \               and self.isValidBSTRecu(root.left, low, root.val) \               and self.isValidBSTRecu(root.right, root.val, high)if __name__ == "__main__":    root, root.left, root.right = TreeNode(2), TreeNode(1), TreeNode(3)    assert Solution().isValidBST(root) == True

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