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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.

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
        }
    }
    
    

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