LeetCode #1339 — MEDIUM

Maximum Product of Splitted Binary Tree

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

The Problem

Problem Statement

Given the root of a binary tree, split the binary tree into two subtrees by removing one edge such that the product of the sums of the subtrees is maximized.

Return the maximum product of the sums of the two subtrees. Since the answer may be too large, return it modulo 10⁹ + 7.

Note that you need to maximize the answer before taking the mod and not after taking it.

Example 1:

Input: root = [1,2,3,4,5,6]
Output: 110
Explanation: Remove the red edge and get 2 binary trees with sum 11 and 10. Their product is 110 (11*10)

Example 2:

Input: root = [1,null,2,3,4,null,null,5,6]
Output: 90
Explanation: Remove the red edge and get 2 binary trees with sum 15 and 6.Their product is 90 (15*6)

Constraints:

The number of nodes in the tree is in the range [2, 5 * 10⁴].
1 <= Node.val <= 10⁴

Step 01

Brute Force Baseline

Problem summary: Given the root of a binary tree, split the binary tree into two subtrees by removing one edge such that the product of the sums of the subtrees is maximized. Return the maximum product of the sums of the two subtrees. Since the answer may be too large, return it modulo 109 + 7. Note that you need to maximize the answer before taking the mod and not after taking it.

Baseline thinking

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

Pattern signal: Tree

Example 1

[1,2,3,4,5,6]

Example 2

[1,null,2,3,4,null,null,5,6]

Core Insight

What unlocks the optimal approach

If we know the sum of a subtree, the answer is max( (total_sum - subtree_sum) * subtree_sum) in each node.

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 #1339: Maximum Product of Splitted 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 long ans;
    private long s;

    public int maxProduct(TreeNode root) {
        final int mod = (int) 1e9 + 7;
        s = sum(root);
        dfs(root);
        return (int) (ans % mod);
    }

    private long dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        long t = root.val + dfs(root.left) + dfs(root.right);
        if (t < s) {
            ans = Math.max(ans, t * (s - t));
        }
        return t;
    }

    private long sum(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return root.val + sum(root.left) + sum(root.right);
    }
}

// Accepted solution for LeetCode #1339: Maximum Product of Splitted Binary Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func maxProduct(root *TreeNode) (ans int) {
	const mod = 1e9 + 7
	var sum func(*TreeNode) int
	sum = func(root *TreeNode) int {
		if root == nil {
			return 0
		}
		return root.Val + sum(root.Left) + sum(root.Right)
	}
	s := sum(root)
	var dfs func(*TreeNode) int
	dfs = func(root *TreeNode) int {
		if root == nil {
			return 0
		}
		t := root.Val + dfs(root.Left) + dfs(root.Right)
		if t < s {
			ans = max(ans, t*(s-t))
		}
		return t
	}
	dfs(root)
	ans %= mod
	return
}

# Accepted solution for LeetCode #1339: Maximum Product of Splitted 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 maxProduct(self, root: Optional[TreeNode]) -> int:
        def sum(root: Optional[TreeNode]) -> int:
            if root is None:
                return 0
            return root.val + sum(root.left) + sum(root.right)

        def dfs(root: Optional[TreeNode]) -> int:
            if root is None:
                return 0
            t = root.val + dfs(root.left) + dfs(root.right)
            nonlocal ans, s
            if t < s:
                ans = max(ans, t * (s - t))
            return t

        mod = 10**9 + 7
        s = sum(root)
        ans = 0
        dfs(root)
        return ans % mod

// Accepted solution for LeetCode #1339: Maximum Product of Splitted 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::rc::Rc;
use std::cell::RefCell;

impl Solution {
    pub fn max_product(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        const MOD: i64 = 1_000_000_007;
        let mut ans: i64 = 0;
        let s = Self::sum(&root);
        Self::dfs(&root, s, &mut ans);
        (ans % MOD) as i32
    }

    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, s: i64, ans: &mut i64) -> i64 {
        if root.is_none() {
            return 0;
        }
        let node = root.as_ref().unwrap().borrow();
        let t = node.val as i64
            + Self::dfs(&node.left, s, ans)
            + Self::dfs(&node.right, s, ans);
        if t < s {
            *ans = (*ans).max(t * (s - t));
        }
        t
    }

    fn sum(root: &Option<Rc<RefCell<TreeNode>>>) -> i64 {
        if root.is_none() {
            return 0;
        }
        let node = root.as_ref().unwrap().borrow();
        node.val as i64 + Self::sum(&node.left) + Self::sum(&node.right)
    }
}

// Accepted solution for LeetCode #1339: Maximum Product of Splitted 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 maxProduct(root: TreeNode | null): number {
    const sum = (root: TreeNode | null): number => {
        if (!root) {
            return 0;
        }
        return root.val + sum(root.left) + sum(root.right);
    };
    const s = sum(root);
    let ans = 0;
    const mod = 1e9 + 7;
    const dfs = (root: TreeNode | null): number => {
        if (!root) {
            return 0;
        }
        const t = root.val + dfs(root.left) + dfs(root.right);
        if (t < s) {
            ans = Math.max(ans, t * (s - t));
        }
        return t;
    };
    dfs(root);
    return ans % mod;
}

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.

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.