comments	difficulty	edit_url
true	简单	https://github.com/doocs/leetcode/edit/main/lcof/%E9%9D%A2%E8%AF%95%E9%A2%9868%20-%20I.%20%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E7%9A%84%E6%9C%80%E8%BF%91%E5%85%AC%E5%85%B1%E7%A5%96%E5%85%88/README.md

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 $[p.val,.. q.val]$ 之间的结点即可。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 = None


class Solution:
    def lowestCommonAncestor(
        self, root: TreeNode, p: TreeNode, q: TreeNode
    ) -> TreeNode:
        while 1:
            if root.val < p.val and root.val < q.val:
                root = root.right
            elif root.val > p.val and root.val > q.val:
                root = root.left
            else:
                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

Python3

# 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: '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 = root
        while 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
    }
}

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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

解法

方法一：一次遍历

Python3

Java

C++

Go

TypeScript

Rust

JavaScript

方法二

Python3

Java

C++

Go

TypeScript

Swift

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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

解法

方法一：一次遍历

Python3

Java

C++

Go

TypeScript

Rust

JavaScript

方法二

Python3

Java

C++

Go

TypeScript

Swift