Maximum Binary Tree II in C++

Suppose we have a root node of a maximum tree: The maximum tree is a tree where every node has a value greater than any other value in its subtree. Suppose we have a method called construct(). This can construct a root from a list A. The construct() method is like −

• If list A is empty, return null.

• Otherwise, let A[i] be the largest element of the list A. Then create a root node with 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[n - 1]]) [n is the length of A]

• Return root.

Note that we were not given A directly, only a root node root = construct(A). Now suppose B is a copy of A with the value val added to it. It is guaranteed that B has unique values. We have to construct(B). If value is 5 and the input tree is like −

The output tree is like −

To solve this, we will follow these steps −

• define one recursive method solve(). This is taking root and val

• If tree is empty, then create a new node with value val, and return that node

• if value of root < val, then

  • temp := new node whose value is val

  • left of temp := root

  • return temp

• right of root := solve(right of root, val)

• return root

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class TreeNode{
   public:
   int val;
   TreeNode *left, *right;
   TreeNode(int data){
      val = data;
      left = NULL;
      right = NULL;
   }
};
void insert(TreeNode **root, int val){
   queue<TreeNode*> q;
   q.push(*root);
   while(q.size()){
      TreeNode *temp = q.front();
      q.pop();
      if(!temp->left){
         if(val != NULL)
            temp->left = new TreeNode(val);
         else
            temp->left = new TreeNode(0);
         return;
      }else{
         q.push(temp->left);
      }
      if(!temp->right){
         if(val != NULL)
            temp->right = new TreeNode(val);
         else
            temp->right = new TreeNode(0);
         return;
      }else{
         q.push(temp->right);
      }
   }
}
TreeNode *make_tree(vector<int> v){
   TreeNode *root = new TreeNode(v[0]);
   for(int i = 1; i<v.size(); i++){
      insert(&root, v[i]);
   }
   return root;
}
void tree_level_trav(TreeNode*root){
   if (root == NULL) return;
   cout << "[";
   queue<TreeNode *> q;
   TreeNode *curr;
   q.push(root);
   q.push(NULL);
   while (q.size() > 1) {
      curr = q.front();
      q.pop();
      if (curr == NULL){
         q.push(NULL);
      }
      else {
         if(curr->left)
            q.push(curr->left);
         if(curr->right)
            q.push(curr->right);
         if(curr == NULL || curr->val == 0){
            cout << "null" << ", ";
         }else{
            cout << curr->val << ", ";
         }
      }
   }
   cout << "]"<<endl;
}
class Solution {
   public:
   TreeNode* insertIntoMaxTree(TreeNode* root, int val) {
      if(!root)return new TreeNode(val);
      if(root->val < val){
         TreeNode* temp = new TreeNode(val);
         temp->left = root;
         return temp;
      }
      root->right = insertIntoMaxTree(root->right, val);
      return root;
   }
};
main(){
   vector<int> v = {4,1,3,NULL,NULL,2};
   TreeNode *root = make_tree(v);
   Solution ob;
   tree_level_trav(ob.insertIntoMaxTree(root, 5));
}

Input

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

Output

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