#998

Medium

auto_awesomeBinary-tree traversal and state tracking

LeetCode Problem Workspace

Maximum Binary Tree II

The Maximum Binary Tree II problem requires building a binary tree by inserting a value into a maximum binary tree while maintaining the tree's properties.

tree binary tree

Problem Statement

You are given the root of a maximum binary tree and an integer value. A maximum binary tree is a binary tree where every node's value is greater than any other value in its subtree. Your task is to insert the given integer into the tree while preserving the maximum binary tree structure.

To insert the integer, follow the rules of the maximum binary tree construction. The new value must be inserted at the correct position where it becomes the root of the largest possible subtree, ensuring that the tree's properties are maintained. The goal is to output the modified tree.

Examples

Example 1

Input: root = [4,1,3,null,null,2], val = 5

Output: [5,4,null,1,3,null,null,2]

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]

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]

a = [2,1,5,3], b = [2,1,5,3,4]

Constraints

The number of nodes in the tree is in the range [1, 100].
1 <= Node.val <= 100
All the values of the tree are unique.
1 <= val <= 100

Solution Approach

Binary Tree Traversal

To solve the problem, perform a tree traversal to find the appropriate spot to insert the new value. The traversal will follow the rules of binary tree insertion, ensuring that the new value is placed at the correct position.

State Tracking

Track the values as you traverse the tree. The challenge is maintaining the correct binary tree properties during insertion, ensuring the new value becomes the root of the largest possible subtree.

Constructing the Tree

After placing the new value, reconstruct the tree to ensure it matches the maximum binary tree properties. This involves adjusting parent-child relationships so that the largest node is always at the root of any subtree.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach, but generally, it is O(N) where N is the number of nodes in the tree. Space complexity may vary depending on the tree structure and recursion depth but will typically be O(H), where H is the height of the tree.

What Interviewers Usually Probe

Ability to perform tree traversal efficiently
Understanding of binary tree properties
Correctly handling state during insertion

Common Pitfalls or Variants

Common pitfalls

Inserting the value at the wrong position in the tree
Not maintaining the binary tree's maximum property after insertion
Failing to update parent-child relationships properly

Follow-up variants

Inserting multiple values into the maximum binary tree
Handling duplicate values in the tree
Optimizing tree construction with iterative methods

FAQ

How do I insert a value into a maximum binary tree?

To insert a value, traverse the tree and place the new value in the correct spot where it becomes the root of the largest possible subtree.

What is a maximum binary tree?

A maximum binary tree is a tree where each node's value is greater than any other value in its subtree.

What is the time complexity of the Maximum Binary Tree II problem?

The time complexity is generally O(N), where N is the number of nodes in the tree.

How does GhostInterview help with tree traversal?

GhostInterview guides you through tree traversal steps, ensuring you correctly identify where to insert the new value in the maximum binary tree.

What are common mistakes in solving Maximum Binary Tree II?

Common mistakes include inserting the value at the wrong position, not maintaining the tree's maximum property, and failing to properly update parent-child relationships.

terminal

Solution

Solution 1: Recursion

If $val$ is the maximum number, then make $val$ the new root node, and $root$ the left subtree of the new root node.

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# 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

Solution 2: Iteration

Search the right subtree, find the node where $curr.val \gt val \gt curr.right.val$, then create a new node $node$, point $node.left$ to $curr.right$, and then point $curr.right$ to $node$.

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# 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

Continue Practicing

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