Maximum Width of Binary Tree in C++

C++Server Side ProgrammingProgramming

Suppose we have a binary tree, we have to define a function to get the maximum width of the given tree. Here the width of a tree is the maximum width among all levels. We will consider the binary tree has the same structure as a full binary tree, but some nodes are null. The width of one level is actually the length between the end-nodes (the leftmost and right most non-null nodes in the level, where the null nodes between the end-nodes are also counted for the length calculation). So if the tree is like −


Then the maximum width is 4, as the nodes of the last layer is [5,3,null,9]

To solve this, we will follow these steps −

  • ans := 1, size := 0

  • define a double ended queue q where we will store (node, value) pair.

  • insert (root, 1) into q

  • while q is not empty

    • size := size of the q

    • define a (node, value) pair curr

    • if size is 1, then (node of the front element, 1) into q, delete element from q

    • while size is not 0

      • curr := front element of q, delete the front element from q

      • if left of the curr node is not null, then

        • create (left of the current node, 2*value of the curr) and insert into q

      • if right of the curr node is not null, then

        • create (right of the current node, 2*value of the curr + 1) and insert into q

      • if size of q > 1, then

        • ans := max of ans, value of the last element in q – value of first element of q + 1

      • size := size – 1

  • return ans

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;
}
class Solution {
   public:
   int widthOfBinaryTree(TreeNode* root) {
      int ans = 0;
      deque < pair <TreeNode*, int> > q;
      q.push_back({root,1});
      ans = 1;
      int size;
      while(!q.empty()){
         size = q.size();
         pair <TreeNode*, int> curr;
         if(size == 1){
            q.push_back({q.front().first, 1});
            q.pop_front();
         }
         while(size--){
            curr = q.front();
            q.pop_front();
            if(curr.first->left){
               q.push_back({curr.first->left, 2 * curr.second});
            }
            if(curr.first->right){
               q.push_back({curr.first->right, 2 * curr.second + 1});
            }
         }
         if(q.size() > 1)
            ans = max(ans, q.back().second - q.front().second + 1);
      }
      return ans;
   }
};
main(){
   vector<int> v = {1,3,2,5,3,NULL,9};
   TreeNode *root = make_tree(v);
   Solution ob;
   cout << (ob.widthOfBinaryTree(root));
}

Input

[1,3,2,5,3,null,9]

Output

4
raja
Published on 04-Mar-2020 07:02:53
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