Leetcode: Lowest Common Ancestor of a Binary Search Tree

Lowest Common Ancestor of a Binary Search Tree



Similar Problems:


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 nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


## Blog link: https://code.dennyzhang.com/lowest-common-ancestor-of-a-binary-search-tree
## Basic Ideas: For BST, get min(p.val, q.val) and max(p.val, q.val)
##              Check from the root node
##              If both are smaller than root.val, move the left sub-tree
##              If both are bigger than root.val, move the right-tree
##              If one smaller and one bigger, the current node is what we want
## Assumption: No duplicate value in the BST
## Complexity: Time O(k), Space O(1). k is the height
# 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
        """
        stack = []
        r = root
        min_val = min(p.val, q.val)
        max_val = max(p.val, q.val)
        while r and (r.val > max_val or r.val < min_val):
            if r.val > max_val:
                r = r.left
            else:
                r = r.right
        return r
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