0235.go

// Source: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree
// Title: Lowest Common Ancestor of a Binary Search Tree
// Difficulty: Easy
// Author: Mu Yang <http://muyang.pro>

////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// Given a binary search tree (BST), find the lowest common ancestor (LCA) node 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).”
//
// Example 1:
//
//   https://assets.leetcode.com/uploads/2018/12/14/binarysearchtree_improved.png
//
//   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.
//
// Example 2:
//
//   https://assets.leetcode.com/uploads/2018/12/14/binarysearchtree_improved.png
//
//   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.
//
// Example 3:
//
//   Input: root = [2,1], p = 2, q = 1
//   Output: 2
//
// Constraints:
//
//   The number of nodes in the tree is in the range [2, 10^5].
//   -10^9 <= Node.val <= 10^9
//   All Node.val are unique.
//   p != q
//   p and q will exist in the BST.
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////

package main

type TreeNode struct {
	Val   int
	Left  *TreeNode
	Right *TreeNode
}

func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
	if p.Val > q.Val {
		p, q = q, p
	}
	return _lowestCommonAncestor(root, p, q)
}

func _lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
	if p.Val > root.Val {
		return _lowestCommonAncestor(root.Right, p, q)
	}

	if q.Val < root.Val {
		return _lowestCommonAncestor(root.Left, p, q)
	}

	return root
}