LeetCode #654 — MEDIUM

Maximum Binary Tree

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

The Problem

Problem Statement

You are given an integer array nums with no duplicates. A maximum binary tree can be built recursively from nums using the following algorithm:

Create a root node whose value is the maximum value in nums.
Recursively build the left subtree on the subarray prefix to the left of the maximum value.
Recursively build the right subtree on the subarray suffix to the right of the maximum value.

Return the maximum binary tree built from nums.

Example 1:

Input: nums = [3,2,1,6,0,5]
Output: [6,3,5,null,2,0,null,null,1]
Explanation: The recursive calls are as follow:
- The largest value in [3,2,1,6,0,5] is 6. Left prefix is [3,2,1] and right suffix is [0,5].
    - The largest value in [3,2,1] is 3. Left prefix is [] and right suffix is [2,1].
        - Empty array, so no child.
        - The largest value in [2,1] is 2. Left prefix is [] and right suffix is [1].
            - Empty array, so no child.
            - Only one element, so child is a node with value 1.
    - The largest value in [0,5] is 5. Left prefix is [0] and right suffix is [].
        - Only one element, so child is a node with value 0.
        - Empty array, so no child.

Example 2:

Input: nums = [3,2,1]
Output: [3,null,2,null,1]

Constraints:

1 <= nums.length <= 1000
0 <= nums[i] <= 1000
All integers in nums are unique.

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums with no duplicates. A maximum binary tree can be built recursively from nums using the following algorithm: Create a root node whose value is the maximum value in nums. Recursively build the left subtree on the subarray prefix to the left of the maximum value. Recursively build the right subtree on the subarray suffix to the right of the maximum value. Return the maximum binary tree built from nums.

Baseline thinking

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

Pattern signal: Array · Stack · Tree

Example 1

[3,2,1,6,0,5]

Example 2

[3,2,1]

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 #654: Maximum 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 {
    private int[] nums;

    public TreeNode constructMaximumBinaryTree(int[] nums) {
        this.nums = nums;
        return dfs(0, nums.length - 1);
    }

    private TreeNode dfs(int l, int r) {
        if (l > r) {
            return null;
        }
        int i = l;
        for (int j = l; j <= r; ++j) {
            if (nums[i] < nums[j]) {
                i = j;
            }
        }
        TreeNode root = new TreeNode(nums[i]);
        root.left = dfs(l, i - 1);
        root.right = dfs(i + 1, r);
        return root;
    }
}

// Accepted solution for LeetCode #654: Maximum Binary Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func constructMaximumBinaryTree(nums []int) *TreeNode {
	var dfs func(l, r int) *TreeNode
	dfs = func(l, r int) *TreeNode {
		if l > r {
			return nil
		}
		i := l
		for j := l; j <= r; j++ {
			if nums[i] < nums[j] {
				i = j
			}
		}
		root := &TreeNode{Val: nums[i]}
		root.Left = dfs(l, i-1)
		root.Right = dfs(i+1, r)
		return root
	}
	return dfs(0, len(nums)-1)
}

# Accepted solution for LeetCode #654: Maximum Binary Tree
# 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 constructMaximumBinaryTree(self, nums: List[int]) -> Optional[TreeNode]:
        def dfs(nums):
            if not nums:
                return None
            val = max(nums)
            i = nums.index(val)
            root = TreeNode(val)
            root.left = dfs(nums[:i])
            root.right = dfs(nums[i + 1 :])
            return root

        return dfs(nums)

// Accepted solution for LeetCode #654: Maximum Binary Tree
// 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 {
    fn construct(nums: &Vec<i32>, start: usize, end: usize) -> Option<Rc<RefCell<TreeNode>>> {
        if start >= end {
            return None;
        }
        let mut idx = 0;
        let mut max_val = -1;
        for i in start..end {
            if nums[i] > max_val {
                idx = i;
                max_val = nums[i];
            }
        }
        Some(Rc::new(RefCell::new(TreeNode {
            val: max_val,
            left: Self::construct(nums, start, idx),
            right: Self::construct(nums, idx + 1, end),
        })))
    }

    pub fn construct_maximum_binary_tree(nums: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
        Self::construct(&nums, 0, nums.len())
    }
}

// Accepted solution for LeetCode #654: Maximum Binary Tree
/**
 * 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 constructMaximumBinaryTree(nums: number[]): TreeNode | null {
    const n = nums.length;
    if (n === 0) {
        return null;
    }
    const [val, i] = nums.reduce((r, v, i) => (r[0] < v ? [v, i] : r), [-1, 0]);
    return new TreeNode(
        val,
        constructMaximumBinaryTree(nums.slice(0, i)),
        constructMaximumBinaryTree(nums.slice(i + 1)),
    );
}

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

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

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.