Description#
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
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://spcdn.pages.dev/leetcode/problems/0236.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree/images/binarytree.png)
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.
Example 2:
![](https://spcdn.pages.dev/leetcode/problems/0236.Lowest%20Common%20Ancestor%20of%20a%20Binary%20Tree/images/binarytree.png)
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
Example 3:
Input: root = [1,2], p = 1, q = 2
Output: 1
Constraints:
- The number of nodes in the tree is in the range
[2, 105]
. -109 <= Node.val <= 109
- All
Node.val
are unique. p != q
p
and q
will exist in the tree.
Solutions#
Solution 1#
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
| # 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)
return root if left and right else (left or right)
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
| /**
* 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;
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
| /**
* 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 || root == p || root == q) return root;
TreeNode* left = lowestCommonAncestor(root->left, p, q);
TreeNode* right = lowestCommonAncestor(root->right, p, q);
if (left && right) return root;
return left ? left : right;
}
};
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
| /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
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
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
| /**
* 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 {
const find = (root: TreeNode | null) => {
if (root == null || root == p || root == q) {
return root;
}
const left = find(root.left);
const right = find(root.right);
if (left != null && right != null) {
return root;
}
if (left != null) {
return left;
}
return right;
};
return find(root);
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
| // 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::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn find(
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.clone();
}
let node = root.as_ref().unwrap().borrow();
let left = Self::find(&node.left, p, q);
let right = Self::find(&node.right, p, q);
match (left.is_some(), right.is_some()) {
(true, false) => left,
(false, true) => right,
(false, false) => None,
(true, true) => root.clone(),
}
}
pub fn lowest_common_ancestor(
root: Option<Rc<RefCell<TreeNode>>>,
p: Option<Rc<RefCell<TreeNode>>>,
q: Option<Rc<RefCell<TreeNode>>>
) -> Option<Rc<RefCell<TreeNode>>> {
Self::find(&root, &p, &q)
}
}
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
| /**
* 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;
};
|