LeetCode #1123 — MEDIUM

Lowest Common Ancestor of Deepest Leaves

Move from brute-force thinking to an efficient approach using hash map strategy.

Solve on LeetCode

The Problem

Problem Statement

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/

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Hash Map · Tree

Example 1

[3,5,1,6,2,0,8,null,null,7,4]

Example 2

[1]

Example 3

[0,1,3,null,2]

Core Insight

What unlocks the optimal approach

Do a postorder traversal.
Then, if both subtrees contain a deepest leaf, you can mark this node as the answer (so far).
The final node marked will be the correct answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1123: Lowest Common Ancestor of Deepest Leaves
/**
 * 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 {
    public TreeNode lcaDeepestLeaves(TreeNode root) {
        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);
    }
}

// Accepted solution for LeetCode #1123: Lowest Common Ancestor of Deepest Leaves
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
type pair struct {
	first  *TreeNode
	second int
}

func lcaDeepestLeaves(root *TreeNode) *TreeNode {
	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
}

# Accepted solution for LeetCode #1123: Lowest Common Ancestor of Deepest Leaves
# 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]

// Accepted solution for LeetCode #1123: Lowest Common Ancestor of Deepest Leaves
struct Solution;
use rustgym_util::*;

trait Postorder {
    fn postorder(&self) -> (usize, TreeLink);
}

impl Postorder for TreeLink {
    fn postorder(&self) -> (usize, TreeLink) {
        use std::cmp::Ordering::*;
        if let Some(node) = self {
            let left = &node.borrow().left;
            let right = &node.borrow().right;
            let (height_l, left_lca) = left.postorder();
            let (height_r, right_lca) = right.postorder();
            match height_l.cmp(&height_r) {
                Less => (height_r + 1, right_lca),
                Greater => (height_l + 1, left_lca),
                Equal => (height_l + 1, self.clone()),
            }
        } else {
            (0, None)
        }
    }
}

impl Solution {
    fn lca_deepest_leaves(root: TreeLink) -> TreeLink {
        let (_, lca) = root.postorder();
        lca
    }
}

#[test]
fn test() {
    let root = tree!(1, tree!(2), tree!(3));
    let res = tree!(1, tree!(2), tree!(3));
    assert_eq!(Solution::lca_deepest_leaves(root), res);
    let root = tree!(1, tree!(2, tree!(4), None), tree!(3));
    let res = tree!(4);
    assert_eq!(Solution::lca_deepest_leaves(root), res);
    let root = tree!(1, tree!(2, tree!(4), tree!(5)), tree!(3));
    let res = tree!(2, tree!(4), tree!(5));
    assert_eq!(Solution::lca_deepest_leaves(root), res);
    let root = tree!(
        1,
        tree!(2, tree!(3, None, tree!(6)), tree!(4, None, tree!(5))),
        None
    );
    let res = tree!(2, tree!(3, None, tree!(6)), tree!(4, None, tree!(5)));
    assert_eq!(Solution::lca_deepest_leaves(root), res);
}

// Accepted solution for LeetCode #1123: Lowest Common Ancestor of Deepest Leaves
/**
 * 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];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.