68 - I. 二叉搜索树的最近公共祖先

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difficulty: 简单
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面试题 68 - I. 二叉搜索树的最近公共祖先

题目描述

给定一个二叉搜索树, 找到该树中两个指定节点的最近公共祖先。

百度百科中最近公共祖先的定义为：“对于有根树 T 的两个结点 p、q，最近公共祖先表示为一个结点 x，满足 x 是 p、q 的祖先且 x 的深度尽可能大（一个节点也可以是它自己的祖先）。”

例如，给定如下二叉搜索树: root = [6,2,8,0,4,7,9,null,null,3,5]

示例 1:

输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
输出: 6 
解释: 节点 2 和节点 8 的最近公共祖先是 6。

示例 2:

输入: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
输出: 2
解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。

说明:

所有节点的值都是唯一的。
p、q 为不同节点且均存在于给定的二叉搜索树中。

注意：本题与主站 235 题相同：https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-search-tree/

解法

方法一：一次遍历

从上到下遍历二叉树，找到第一个值位于 $[p . v a l, .. q . v a l]$ 之间的结点即可【核心】。既可以用迭代实现，也可以用递归实现。

时间复杂度 $O (n)$ ，其中 $n$ 是二叉树的结点数。空间复杂度方面，迭代实现的空间复杂度为 $O (1)$ ，递归实现的空间复杂度为 $O (n)$ 。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def lowestCommonAncestor(self, root: TreeNode, p: TreeNode, q: TreeNode) -> TreeNode:while 1:if root.val < p.val and root.val < q.val:   # 说明p，q节点在右子树，因此公共祖先节点也应该向右边找root = root.rightelif root.val > p.val and root.val > q.val: # 反之，亦然root = root.leftelse:return root

Java

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {while (true) {if (root.val < p.val && root.val < q.val) {root = root.right;} else if (root.val > p.val && root.val > q.val) {root = root.left;} else {return root;}}}
}

C++

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {if (root->val < p->val && root->val < q->val) {return lowestCommonAncestor(root->right, p, q);}if (root->val > p->val && root->val > q->val) {return lowestCommonAncestor(root->left, p, q);}return root;}
};

Go

/*** Definition for a binary tree node.* type TreeNode struct {*     Val   int*     Left  *TreeNode*     Right *TreeNode* }*/func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {if root.Val < p.Val && root.Val < q.Val {return lowestCommonAncestor(root.Right, p, q)}if root.Val > p.Val && root.Val > q.Val {return lowestCommonAncestor(root.Left, p, q)}return root
}

TypeScript

/*** Definition for a binary tree node.* class TreeNode {*     val: number*     left: TreeNode | null*     right: TreeNode | null*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {*         this.val = (val===undefined ? 0 : val)*         this.left = (left===undefined ? null : left)*         this.right = (right===undefined ? null : right)*     }* }*/
function lowestCommonAncestor(root: TreeNode | null,p: TreeNode | null,q: TreeNode | null,
): TreeNode | null {if (root == null) {return root;}if (root.val > p.val && root.val > q.val) {return lowestCommonAncestor(root.left, p, q);}if (root.val < p.val && root.val < q.val) {return lowestCommonAncestor(root.right, p, q);}return root;
}

Rust

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::cmp::Ordering;
use std::rc::Rc;
impl Solution {pub fn lowest_common_ancestor(mut root: Option<Rc<RefCell<TreeNode>>>,p: Option<Rc<RefCell<TreeNode>>>,q: Option<Rc<RefCell<TreeNode>>>,) -> Option<Rc<RefCell<TreeNode>>> {let p = p.unwrap().borrow().val;let q = q.unwrap().borrow().val;loop {let mut cur = root.as_ref().unwrap().borrow().val;match (cur.cmp(&p), cur.cmp(&q)) {(Ordering::Less, Ordering::Less) => {root = root.unwrap().borrow().right.clone();}(Ordering::Greater, Ordering::Greater) => {root = root.unwrap().borrow().left.clone();}(_, _) => {break root;}}}}
}

JavaScript

/*** Definition for a binary tree node.* function TreeNode(val) {*     this.val = val;*     this.left = this.right = null;* }*/
/*** @param {TreeNode} root* @param {TreeNode} p* @param {TreeNode} q* @return {TreeNode}*/
var lowestCommonAncestor = function (root, p, q) {if (root.val < p.val && root.val < q.val) {return lowestCommonAncestor(root.right, p, q);} else if (root.val > p.val && root.val > q.val) {return lowestCommonAncestor(root.left, p, q);}return root;
};

方法二：递归

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':if root.val < p.val and root.val < q.val:return self.lowestCommonAncestor(root.right, p, q)if root.val > p.val and root.val > q.val:return self.lowestCommonAncestor(root.left, p, q)return root

Java

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if (root.val < p.val && root.val < q.val) {return lowestCommonAncestor(root.right, p, q);}if (root.val > p.val && root.val > q.val) {return lowestCommonAncestor(root.left, p, q);}return root;}
}

C++

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}* };*/
class Solution {
public:TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {while (1) {if (root->val < p->val && root->val < q->val) {root = root->right;} else if (root->val > p->val && root->val > q->val) {root = root->left;} else {return root;}}}
};

Go

/*** Definition for a binary tree node.* type TreeNode struct {*     Val   int*     Left  *TreeNode*     Right *TreeNode* }*/func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {for {if root.Val < p.Val && root.Val < q.Val {root = root.Right} else if root.Val > p.Val && root.Val > q.Val {root = root.Left} else {return root}}
}

TypeScript

/*** Definition for a binary tree node.* class TreeNode {*     val: number*     left: TreeNode | null*     right: TreeNode | null*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {*         this.val = (val===undefined ? 0 : val)*         this.left = (left===undefined ? null : left)*         this.right = (right===undefined ? null : right)*     }* }*/
function lowestCommonAncestor(root: TreeNode | null,p: TreeNode | null,q: TreeNode | null,
): TreeNode | null {if (root == null) {return root;}while (true) {if (root.val > p.val && root.val > q.val) {root = root.left;} else if (root.val < p.val && root.val < q.val) {root = root.right;} else {return root;}}
}

Swift

/*** Definition for a binary tree node.* public class TreeNode {*     public var val: Int*     public var left: TreeNode?*     public var right: TreeNode?*     public init(_ val: Int) {*         self.val = val*         self.left = nil*         self.right = nil*     }* }*/class Solution {func lowestCommonAncestor(_ root: TreeNode?, _ p: TreeNode?, _ q: TreeNode?) -> TreeNode? {guard let p = p, let q = q else {return nil}var node = rootwhile let current = node {if current.val < p.val && current.val < q.val {node = current.right} else if current.val > p.val && current.val > q.val {node = current.left} else {return current}}return nil}
}