Level Order Traversal of Binary Tree in C++

Suppose we have a binary tree. We have to traverse this tree using level order traversal fashion. So if the tree is like

The traversal sequence will be like: [1,2,3,5,4]

To solve this, we will follow these steps −

• define queue que to store nodes

• insert root into the que.

• while que is not empty, do

  • item := item present at front position of queue

  • print the value of item

  • if left of the item is not null, then insert left of item into que

  • if right of the item is not null, then insert right of item into que

  • delete front element from que

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<int> v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << v[i] << ", ";
   }
   cout << "]"<<endl;
}
class TreeNode{
   public:
      int val;
      TreeNode *left, *right;
      TreeNode(int data){
         val = data;
         left = right = NULL;
      }
};
class Solution {
   public:
   vector<int> solve(TreeNode* root) {
      if(!root)
         return {};
      vector <int> ret;
      queue <TreeNode*> q;
      q.push(root);
      while(!q.empty()){
         TreeNode* node = q.front();
         q.pop();
         ret.push_back(node->val);
         if(node->left){
            q.push(node->left);
         }
         if(node->right){
            q.push(node->right);
         }
      }
      return ret;
   }
};
main(){
   TreeNode *root = new TreeNode(1);
   root->left = new TreeNode(2);
   root->right = new TreeNode(3);
   root->left->right = new TreeNode(5);
   root->right->right = new TreeNode(4);
   Solution ob;
   print_vector(ob.solve(root));
}

Input

TreeNode *root = new TreeNode(1);
root->left = new TreeNode(2);
root->right = new TreeNode(3);
root->left->right = new TreeNode(5);
root->right->right = new TreeNode(4);

Output

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