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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.

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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

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/**
 * 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;
    }
}

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/**
 * 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;
    }
};

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/**
 * 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
}

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/**
 * 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;
}

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// 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
    }
}

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

struct TreeNode* insertIntoMaxTree(struct TreeNode* root, int val) {
    if (!root || root->val < val) {
        struct TreeNode* res = (struct TreeNode*) malloc(sizeof(struct TreeNode));
        res->val = val;
        res->left = root;
        res->right = NULL;
        return res;
    }
    root->right = 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 \(node\).

Finally, return \(root\).

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

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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.val < val:
            return TreeNode(val, root)
        curr = root
        node = TreeNode(val)
        while curr.right and curr.right.val > val:
            curr = curr.right
        node.left = curr.right
        curr.right = node
        return root

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/**
 * 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.val < val) {
            return new TreeNode(val, root, null);
        }
        TreeNode curr = root;
        TreeNode node = new TreeNode(val);
        while (curr.right != null && curr.right.val > val) {
            curr = curr.right;
        }
        node.left = curr.right;
        curr.right = node;
        return root;
    }
}

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/**
 * 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->val < val) return new TreeNode(val, root, nullptr);
        TreeNode* curr = root;
        TreeNode* node = new TreeNode(val);
        while (curr->right && curr->right->val > val) curr = curr->right;
        node->left = curr->right;
        curr->right = node;
        return root;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func insertIntoMaxTree(root *TreeNode, val int) *TreeNode {
    if root.Val < val {
        return &TreeNode{val, root, nil}
    }
    node := &TreeNode{Val: val}
    curr := root
    for curr.Right != nil && curr.Right.Val > val {
        curr = curr.Right
    }
    node.Left = curr.Right
    curr.Right = node
    return root
}

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/**
 * 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.val < val) {
        return new TreeNode(val, root);
    }
    const node = new TreeNode(val);
    let curr = root;
    while (curr.right && curr.right.val > val) {
        curr = curr.right;
    }
    node.left = curr.right;
    curr.right = node;
    return root;
}

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