| comments | difficulty | edit_url |
|---|---|---|
true |
简单 |
给定一个二叉树, 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，给定如下二叉树: root = [3,5,1,6,2,0,8,null,null,7,4]
示例 1:
输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1 输出: 3 解释: 节点 5 和节点 1 的最近公共祖先是节点 3。
示例 2:
输入: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4 输出: 5 解释: 节点 5 和节点 4 的最近公共祖先是节点 5。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树中。
注意：本题与主站 236 题相同：https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-tree/
根据“最近公共祖先”的定义，若
- 如果
$p$ 和$q$ 分别是$root$ 的左右节点，那么$root$ 就是我们要找的最近公共祖先； - 如果
$p$ 和$q$ 都是$root$ 的左节点，那么返回$lowestCommonAncestor(root.left, p, q)$ ； - 如果
$p$ 和$q$ 都是$root$ 的右节点，那么返回$lowestCommonAncestor(root.right, p, q)$ 。
边界条件讨论：
- 如果
$root$ 为null，则说明我们已经找到最底了，返回null表示没找到； - 如果
$root$ 与$p$ 相等或者与$q$ 相等，则返回$root$ ； - 如果左子树没找到，递归函数返回
null，证明$p$ 和$q$ 同在$root$ 的右侧，那么最终的公共祖先就是右子树找到的结点； - 如果右子树没找到，递归函数返回
null，证明$p$ 和$q$ 同在$root$ 的左侧，那么最终的公共祖先就是左子树找到的结点。
时间复杂度
# 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 is None or root == p or root == q:
return root
left = self.lowestCommonAncestor(root.left, p, q)
right = self.lowestCommonAncestor(root.right, p, q)
if left is None:
return right
if right is None:
return left
return root/**
* 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 == null || root == p || root == q) return root;
TreeNode left = lowestCommonAncestor(root.left, p, q);
TreeNode right = lowestCommonAncestor(root.right, p, q);
if (left == null) return right;
if (right == null) return left;
return root;
}
}/**
* 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) {
// 如果找到val，层层向上传递该root
if (nullptr == root || p->val == root->val || q->val == root->val) {
return root;
}
TreeNode* left = lowestCommonAncestor(root->left, p, q);
TreeNode* right = lowestCommonAncestor(root->right, p, q);
if (left != nullptr && right != nullptr) {
// 如果两边都可以找到
return root;
} else if (left == nullptr) {
// 如果左边没有找到，则直接返回右边内容
return right;
} else {
return left;
}
}
};func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode {
if root == nil || root == p || root == q {
return root
}
left := lowestCommonAncestor(root.Left, p, q)
right := lowestCommonAncestor(root.Right, p, q)
if left == nil {
return right
}
if right == nil {
return left
}
return root
}/**
* 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 || root === p || root === q) {
return root;
}
const left = lowestCommonAncestor(root.left, p, q);
const right = lowestCommonAncestor(root.right, p, q);
if (left == null && right == null) {
return null;
}
if (left == null) {
return right;
}
if (right == null) {
return left;
}
return root;
}// 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::rc::Rc;
impl Solution {
pub fn lowest_common_ancestor(
root: Option<Rc<RefCell<TreeNode>>>,
p: Option<Rc<RefCell<TreeNode>>>,
q: Option<Rc<RefCell<TreeNode>>>,
) -> Option<Rc<RefCell<TreeNode>>> {
if root.is_none() || root == p || root == q {
return root;
}
let left = Self::lowest_common_ancestor(
root.as_ref().unwrap().borrow_mut().left.take(),
p.clone(),
q.clone(),
);
let right = Self::lowest_common_ancestor(
root.as_ref().unwrap().borrow_mut().right.take(),
p.clone(),
q.clone(),
);
match (left.is_none(), right.is_none()) {
(true, false) => right,
(false, true) => left,
(false, false) => root,
(true, true) => None,
}
}
}/**
* 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 || root == p || root == q) return root;
const left = lowestCommonAncestor(root.left, p, q);
const right = lowestCommonAncestor(root.right, p, q);
if (!left) return right;
if (!right) return left;
return root;
};/* public class TreeNode {
* public var val: Int
* public var left: TreeNode?
* public var right: TreeNode?
* public init() { self.val = 0; self.left = nil; self.right = nil; }
* public init(_ val: Int) { self.val = val; self.left = nil; self.right = nil; }
* public init(_ val: Int, _ left: TreeNode?, _ right: TreeNode?) {
* self.val = val
* self.left = left
* self.right = right
* }
* }
*/
class Solution {
func lowestCommonAncestor(_ root: TreeNode?, _ p: TreeNode, _ q: TreeNode) -> TreeNode? {
if root == nil || root === p || root === q {
return root
}
let left = lowestCommonAncestor(root?.left, p, q)
let right = lowestCommonAncestor(root?.right, p, q)
if let _ = left, let _ = right {
return root
}
return left ?? right
}
}