Leetcode: Lowest Common Ancestor of a Binary Search Tree

// https://code.dennyzhang.com/lowest-common-ancestor-of-a-binary-search-tree
// Basic Ideas:
//
// Complexity: Time O(log(n)), Space O(1)
/**
 * Definition for TreeNode.
 * type TreeNode struct {
 *     Val int
 *     Left *ListNode
 *     Right *ListNode
 * }
 */
func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
    if root == nil {
        return root
    }
    if p.Val > q.Val {
        p, q = q, p
    }
    res := root
    for res != nil {
        if res == p || res == q {
            return res
        }
        if res.Val > p.Val && res.Val < q.Val {
            return res
        }
        if res.Val > q.Val {
            res = res.Left
        } else {
            res = res.Right
        }
    }
    return res
}

// https://code.dennyzhang.com/lowest-common-ancestor-of-a-binary-search-tree
// Basic Ideas:
//
// Complexity: Time O(log(n)), Space O(1)
/**
 * Definition for TreeNode.
 * type TreeNode struct {
 *     Val int
 *     Left *ListNode
 *     Right *ListNode
 * }
 */
func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
    res := root
    for res != nil {
        if res.Val > p.Val && res.Val > q.Val {
            res = res.Left
        } else {
            if res.Val < p.Val && res.Val < q.Val {
                res = res.Right
            } else {
                return res
            }
        }
    }
    return res
}

// https://code.dennyzhang.com/lowest-common-ancestor-of-a-binary-search-tree
// Basic Ideas: recursive
//
// Complexity: Time O(log(n)), Space O(1)
/**
 * Definition for TreeNode.
 * type TreeNode struct {
 *     Val int
 *     Left *ListNode
 *     Right *ListNode
 * }
 */
func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
    if root == nil || root == p || root == q {
        return root
    }
    if p.Val > q.Val {
        p, q = q, p
    }
    if root.Val > p.Val && root.Val < q.Val {
        return root
    } else {
        if root.Val > q.Val {
            return lowestCommonAncestor(root.Left, p, q)
        } else {
            return lowestCommonAncestor(root.Right, p, q)
        }
    }
}

## 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(log(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
        """
        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