235.Lowest Common Ancestor of a Binary Search Tree

Tags: [tree], [binary_tree], [binary_search_tree], [BST], [recursion], [extend_function_meaning]

Com: {fb}

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

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

        _______6______
       /              \
    ___2__          ___8__
   /      \        /      \
   0      _4       7       9
         /  \
         3   5

For example, the lowest common ancestor (LCA) of nodes2and8is6. Another example is LCA of nodes2and4is2, since a node can be a descendant of itself according to the LCA definition.

Another solution: Using the BST property

# 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
        """
        # base case
        if not root:
            return None
        if min(p.val, q.val) <= root.val <= max(p.val, q.val):
            return root

        left = self.lowestCommonAncestor(root.left, p, q)
        right = self.lowestCommonAncestor(root.right, p, q)

        return left if left else right

Note:

• Time complexity = O(n), n is the number of nodes of the given tree, because each node we only visited once.

Solution: Extend Function Meaning

# 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
        """
        # base case
        if root in (None, p, q):
            return root

        left = self.lowestCommonAncestor(root.left, p, q)
        right = self.lowestCommonAncestor(root.right, p, q)

        if left and right:
            return root

        return left if left else right

Note:

• Time complexity = O(n), n is the number of nodes of the given tree.