Leetcode 235. Lowest Common Ancestor of a Binary Search Tree

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/

Because BST is a knid of binary tree, so if we use the result of binary tree, it is also correct but not that fast.

If we use BST's feature, the thing is clear: when the three nodes are on the same tree, there should be
s <= root.val <= b. If not, the two nodes must be on left  or right subtree, then use the value to decide go left or right.

Solution:

# T:O(n) S:O(1)
class Solution(object):
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        s, b = sorted([p.val, q.val])
        while not s <= root.val <= b:
            # Keep searching since root is outside of [s, b].
            root = root.left if s <= root.val else root.right
        # s <= root.val <= b.
        return root

Run Time: 132 ms