# 998. Maximum Binary Tree II

## Description

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).

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:

• 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

## Solutions

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.

If $val$ is not the maximum number, since $val$ is the last appended number, it must be on the right side of $root$. Therefore, we can insert $val$ as a new node into the right subtree of $root$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the tree.

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 $node$.

Finally, return $root$.

The time complexity is $O(n)$, where $n$ is the number of nodes in the tree. The space complexity is $O(1)$.

• /**
* 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;
}
}

• /**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode() : val(0), left(nullptr), right(nullptr) {}
*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* insertIntoMaxTree(TreeNode* root, int val) {
if (!root || root->val < val) return new TreeNode(val, root, nullptr);
root->right = insertIntoMaxTree(root->right, val);
return root;
}
};

• # 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


• /**
* 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
}

• /**
* 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;
}


• // 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 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
}
}