LeetCode #865 — MEDIUM

Smallest Subtree with all the Deepest Nodes

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, the depth of each node is the shortest distance to the root.

Return the smallest subtree such that it contains all the deepest nodes in the original tree.

A node is called the deepest if it has the largest depth possible among any node in the entire tree.

The subtree of a node is a tree consisting of that node, plus the set of all descendants of that node.

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 nodes of the tree.
Notice that nodes 5, 3 and 2 contain the deepest nodes in the tree but node 2 is the smallest subtree among them, so we return it.

Example 2:

Input: root = [1]
Output: [1]
Explanation: The root is the deepest node in the tree.

Example 3:

Input: root = [0,1,3,null,2]
Output: [2]
Explanation: The deepest node in the tree is 2, the valid subtrees are the subtrees of nodes 2, 1 and 0 but the subtree of node 2 is the smallest.

Constraints:

The number of nodes in the tree will be in the range [1, 500].
0 <= Node.val <= 500
The values of the nodes in the tree are unique.

Note: This question is the same as 1123: https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/

Step 01

Brute Force Baseline

Problem summary: Given the root of a binary tree, the depth of each node is the shortest distance to the root. Return the smallest subtree such that it contains all the deepest nodes in the original tree. A node is called the deepest if it has the largest depth possible among any node in the entire tree. The subtree of a node is a tree consisting of that node, plus the set of all descendants of that node.

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]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #865: Smallest Subtree with all the Deepest Nodes
/**
 * 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 subtreeWithAllDeepest(TreeNode root) {
        return dfs(root).getKey();
    }

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

// Accepted solution for LeetCode #865: Smallest Subtree with all the Deepest Nodes
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func subtreeWithAllDeepest(root *TreeNode) *TreeNode {
	type pair struct {
		node  *TreeNode
		depth int
	}
	var dfs func(*TreeNode) pair
	dfs = func(root *TreeNode) pair {
		if root == nil {
			return pair{nil, 0}
		}
		l, r := dfs(root.Left), dfs(root.Right)
		ld, rd := l.depth, r.depth
		if ld > rd {
			return pair{l.node, ld + 1}
		}
		if ld < rd {
			return pair{r.node, rd + 1}
		}
		return pair{root, ld + 1}
	}
	return dfs(root).node
}

# Accepted solution for LeetCode #865: Smallest Subtree with all the Deepest Nodes
# 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 subtreeWithAllDeepest(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
        def dfs(root: Optional[TreeNode]) -> Tuple[Optional[TreeNode], int]:
            if root is None:
                return None, 0
            l, ld = dfs(root.left)
            r, rd = dfs(root.right)
            if ld > rd:
                return l, ld + 1
            if ld < rd:
                return r, rd + 1
            return root, ld + 1

        return dfs(root)[0]

// Accepted solution for LeetCode #865: Smallest Subtree with all the Deepest Nodes
// 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 subtree_with_all_deepest(
        root: Option<Rc<RefCell<TreeNode>>>,
    ) -> Option<Rc<RefCell<TreeNode>>> {
        fn dfs(
            root: Option<Rc<RefCell<TreeNode>>>,
        ) -> (Option<Rc<RefCell<TreeNode>>>, i32) {
            if let Some(node) = root {
                let left = node.borrow().left.clone();
                let right = node.borrow().right.clone();

                let (l, ld) = dfs(left);
                let (r, rd) = dfs(right);

                if ld > rd {
                    (l, ld + 1)
                } else if ld < rd {
                    (r, rd + 1)
                } else {
                    (Some(node), ld + 1)
                }
            } else {
                (None, 0)
            }
        }

        dfs(root).0
    }
}

// Accepted solution for LeetCode #865: Smallest Subtree with all the Deepest Nodes
/**
 * 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 subtreeWithAllDeepest(root: TreeNode | null): TreeNode | null {
    const dfs = (root: TreeNode | null): [TreeNode, number] => {
        if (!root) {
            return [null, 0];
        }
        const [l, ld] = dfs(root.left);
        const [r, rd] = dfs(root.right);
        if (ld > rd) {
            return [l, ld + 1];
        }
        if (ld < rd) {
            return [r, rd + 1];
        }
        return [root, ld + 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.