LeetCode #235 Lowest Common Ancestor of a Binary Search Tree

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Problem

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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Examples

Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]

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     _______6______
    /              \
 ___2__          ___8__
/      \        /      \
0      _4       7       9
      /  \
      3   5
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Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.
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Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself 
             according to the LCA definition.

Solution1

Method: Recursive

Time Complexity:

Space Complexity:

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        if p.val > q.val:
            return self.lowestCommonAncestor(root, q, p)
        if root.val < p.val:
            return self.lowestCommonAncestor(root.right, p, q)
        elif root.val > q.val:
            return self.lowestCommonAncestor(root.left, p, q)
        else:
            return root

Solution2

Method: Iterative

Time Complexity

Space Complexity

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        if p.val > q.val:
            return self.lowestCommonAncestor(root, q, p)

        node = root

        while node.val < p.val or node.val > q.val:
            if node.val < p.val:
                node = node.right
            else:
                node = node.left
        return node