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.
Move from brute-force thinking to an efficient approach using tree strategy.
A maximum tree is a tree where every node has a value greater than any other value in its subtree.
You are given the root of a maximum binary tree and an integer val.
Just as in the previous problem, the given tree was constructed from a list a (root = Construct(a)) recursively with the following Construct(a) routine:
a is empty, return null.a[i] be the largest element of a. Create a root node with the value a[i].root will be Construct([a[0], a[1], ..., a[i - 1]]).root will be Construct([a[i + 1], a[i + 2], ..., a[a.length - 1]]).root.Note that we were not given a directly, only a root node root = Construct(a).
Suppose b is a copy of a with the value val appended to it. It is guaranteed that b has unique values.
Return Construct(b).
Example 1:
Input: root = [4,1,3,null,null,2], val = 5 Output: [5,4,null,1,3,null,null,2] Explanation: a = [1,4,2,3], b = [1,4,2,3,5]
Example 2:
Input: root = [5,2,4,null,1], val = 3 Output: [5,2,4,null,1,null,3] Explanation: a = [2,1,5,4], b = [2,1,5,4,3]
Example 3:
Input: root = [5,2,3,null,1], val = 4 Output: [5,2,4,null,1,3] Explanation: a = [2,1,5,3], b = [2,1,5,3,4]
Constraints:
[1, 100].1 <= Node.val <= 1001 <= val <= 100Problem summary: A maximum tree is a tree where every node has a value greater than any other value in its subtree. You are given the root of a maximum binary tree and an integer val. Just as in the previous problem, the given tree was constructed from a list a (root = Construct(a)) recursively with the following Construct(a) routine: If a is empty, return null. Otherwise, let a[i] be the largest element of a. Create a root node with the value a[i]. The left child of root will be Construct([a[0], a[1], ..., a[i - 1]]). The right child of root will be Construct([a[i + 1], a[i + 2], ..., a[a.length - 1]]). Return root. Note that we were not given a directly, only a root node root = Construct(a). Suppose b is a copy of a with the value val appended to it. It is guaranteed that b has unique values. Return Construct(b).
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Tree
[4,1,3,null,null,2] 5
[5,2,4,null,1] 3
[5,2,3,null,1] 4
maximum-binary-tree)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #998: Maximum Binary Tree II
/**
* 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 insertIntoMaxTree(TreeNode root, int val) {
if (root == null || root.val < val) {
return new TreeNode(val, root, null);
}
root.right = insertIntoMaxTree(root.right, val);
return root;
}
}
// Accepted solution for LeetCode #998: Maximum Binary Tree II
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func insertIntoMaxTree(root *TreeNode, val int) *TreeNode {
if root == nil || root.Val < val {
return &TreeNode{val, root, nil}
}
root.Right = insertIntoMaxTree(root.Right, val)
return root
}
# Accepted solution for LeetCode #998: Maximum Binary Tree II
# 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 insertIntoMaxTree(
self, root: Optional[TreeNode], val: int
) -> Optional[TreeNode]:
if root is None or root.val < val:
return TreeNode(val, root)
root.right = self.insertIntoMaxTree(root.right, val)
return root
// Accepted solution for LeetCode #998: Maximum Binary Tree II
// 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 insert_into_max_tree(
mut root: Option<Rc<RefCell<TreeNode>>>,
val: i32,
) -> Option<Rc<RefCell<TreeNode>>> {
if root.is_none() || root.as_ref().unwrap().as_ref().borrow().val < val {
return Some(Rc::new(RefCell::new(TreeNode {
val,
left: root.take(),
right: None,
})));
}
{
let mut root = root.as_ref().unwrap().as_ref().borrow_mut();
root.right = Self::insert_into_max_tree(root.right.take(), val);
}
root
}
}
// Accepted solution for LeetCode #998: Maximum Binary Tree II
/**
* 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 insertIntoMaxTree(root: TreeNode | null, val: number): TreeNode | null {
if (!root || root.val < val) {
return new TreeNode(val, root);
}
root.right = insertIntoMaxTree(root.right, val);
return root;
}
Use this to step through a reusable interview workflow for this problem.
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.
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.
Review these before coding to avoid predictable interview regressions.
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.