235_Lowest Common Ancestor of a Binary Search Tree

235. Lowest Common Ancestor of a Binary Search Tree

Level: easy

Tag: tree

Question

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).”

Example 1

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

LCA of nodes 2 and 8 is 6. 
LCA of nodes 2 and 4 is 2

Idea

Search from up to down, for each node, there are three possibilities:

• node value is larger than p and q, then move node to the left child

• node value is smaller than p and q, then move node to the right child

• node value is between p and q, then this node is the LCA.

Solution: iterative

Time: $O(n)$

Space: $O(1)$

# 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
        """

        while root:
            if p.val < root.val > q.val:
                # move to left child
                root = root.left
            elif p.val > root.val < q.val:
                # move to right child
                root = root.right
            else:
                return root

Solution: recursive

Time: $O(n)$

Space: $O(1)$

# 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 not root:
            return root
        if p.val > q.val:
            return self.lowestCommonAncestor(root, q, p)
        if root.val >= p.val and root.val <= q.val:
            return root
        if root.val < p.val:
            return self.lowestCommonAncestor(root.right, p, q)
        if root.val > q.val:
            return self.lowestCommonAncestor(root.left, p, q)