Sandbox

Problem Solution Time Complexity O(N) Space Complexity O(1) # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def lowestCommonAncestor(self, root, p, q): """ :type root: TreeNode :type p: TreeNode :type q: TreeNode :rtype: TreeNode """ # If p and q are on the same subtree of the cu..

Problem Recursive Solution Use inorder traversal to go through the BST. If the value of current node is smaller than or equal to the value of previous node, the BST is not valid. When using recursion, verify the left side of the tree first. After reaching the very left node, move up to the root while checking the right nodes of each node on the way. Then check the right side of the tree. I tried..