# 1123. Lowest Common Ancestor of Deepest Leaves

## Description

Given the root of a binary tree, return the lowest common ancestor of its deepest leaves.

Recall that:

• The node of a binary tree is a leaf if and only if it has no children
• The depth of the root of the tree is 0. if the depth of a node is d, the depth of each of its children is d + 1.
• The lowest common ancestor of a set S of nodes, is the node A with the largest depth such that every node in S is in the subtree with root A.

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4]
Output: [2,7,4]
Explanation: We return the node with value 2, colored in yellow in the diagram.
The nodes coloured in blue are the deepest leaf-nodes of the tree.
Note that nodes 6, 0, and 8 are also leaf nodes, but the depth of them is 2, but the depth of nodes 7 and 4 is 3.

Example 2:

Input: root = [1]
Output: [1]
Explanation: The root is the deepest node in the tree, and it's the lca of itself.


Example 3:

Input: root = [0,1,3,null,2]
Output: [2]
Explanation: The deepest leaf node in the tree is 2, the lca of one node is itself.


Constraints:

• The number of nodes in the tree will be in the range [1, 1000].
• 0 <= Node.val <= 1000
• The values of the nodes in the tree are unique.

Note: This question is the same as 865: https://leetcode.com/problems/smallest-subtree-with-all-the-deepest-nodes/

## Solutions

Solution 1: DFS

We design a function dfs(root) that returns a tuple (l, d), where l is the deepest common ancestor of node root, and d is the depth of node root. The execution logic of the function dfs(root) is as follows:

• If root is null, return the tuple (None, 0);
• Otherwise, we recursively call dfs(root.left) and dfs(root.right), obtaining tuples (l, d1) and (r, d2). If d1 > d2, the deepest common ancestor of root is l, and the depth is d1 + 1; if d1 < d2, the deepest common ancestor of root is r, and the depth is d2 + 1; if d1 = d2, the deepest common ancestor of root is root, and the depth is d1 + 1.

In the main function, we call dfs(root) and return the first element of its return value to get the deepest common ancestor node.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
return dfs(root).getKey();
}

private Pair<TreeNode, Integer> dfs(TreeNode root) {
if (root == null) {
return new Pair<>(null, 0);
}
Pair<TreeNode, Integer> l = dfs(root.left);
Pair<TreeNode, Integer> r = dfs(root.right);
int d1 = l.getValue(), d2 = r.getValue();
if (d1 > d2) {
return new Pair<>(l.getKey(), d1 + 1);
}
if (d1 < d2) {
return new Pair<>(r.getKey(), d2 + 1);
}
return new Pair<>(root, d1 + 1);
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
return dfs(root).first;
}

pair<TreeNode*, int> dfs(TreeNode* root) {
if (!root) {
return {nullptr, 0};
}
auto [l, d1] = dfs(root->left);
auto [r, d2] = dfs(root->right);
if (d1 > d2) {
return {l, d1 + 1};
}
if (d1 < d2) {
return {r, d2 + 1};
}
return {root, d1 + 1};
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def lcaDeepestLeaves(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
def dfs(root):
if root is None:
return None, 0
l, d1 = dfs(root.left)
r, d2 = dfs(root.right)
if d1 > d2:
return l, d1 + 1
if d1 < d2:
return r, d2 + 1
return root, d1 + 1

return dfs(root)[0]


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
type pair struct {
first  *TreeNode
second int
}

var dfs func(root *TreeNode) pair
dfs = func(root *TreeNode) pair {
if root == nil {
return pair{nil, 0}
}
l, r := dfs(root.Left), dfs(root.Right)
d1, d2 := l.second, r.second
if d1 > d2 {
return pair{l.first, d1 + 1}
}
if d1 < d2 {
return pair{r.first, d2 + 1}
}
return pair{root, d1 + 1}
}
return dfs(root).first
}

• /**
* 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 lcaDeepestLeaves(root: TreeNode | null): TreeNode | null {
const dfs = (root: TreeNode | null): [TreeNode | null, number] => {
if (root === null) {
return [null, 0];
}
const [l, d1] = dfs(root.left);
const [r, d2] = dfs(root.right);
if (d1 > d2) {
return [l, d1 + 1];
}
if (d1 < d2) {
return [r, d2 + 1];
}
return [root, d1 + 1];
};
return dfs(root)[0];
}