{"post":"

Breadth-First Search Traversal of a Tree

For breadth-first traversal, we will use a queue. We will first push a node to it and for each node in front, we’ll check if its left and right child exist if yes then push it to the queue, and in the next iteration pop front and print it.

#include<bits/stdc++.h>\r\nusing namespace std;\nstruct Node{\r\nint data;\r\nNode* left;\r\nNode* right;\r\nNode(int data){\r\nthis->data = data;\r\nleft = right = NULL;\r\n}\r\n};
\nvoid bfs(Node* root){\r\nif(root == NULL) return;\r\nqueue<Node*> q;\r\nq.push(root);\r\nwhile(!q.empty()){\r\nint size = q.size();
\n  \tfor(int i = 0; i &lt; size; i++){\r\n    \tNode* curr = q.front();\r\n      \tq.pop();\r\n      \tcout&lt;&lt;curr-&gt;data&lt;&lt;\" \";\r\n      \tif(curr-&gt;left) q.push(curr-&gt;left);\r\n        if(curr-&gt;right) q.push(curr-&gt;right);\r\n    }\r\n}\n
\n}
\nint main() {\r\n/*\r\n1\r\n/    
\n2      3\r\n/   \\   /  
\n4     5 6    7\r\n/\r\nstruct Node root = new Node(1);\r\nroot->left = new Node(2);\r\nroot->left->left = new Node(4);\r\nroot->left->right = new Node(5);\r\nroot->right = new Node(3);\r\nroot->right->left = new Node(6);\r\nroot->right->right = new Node(7);\r\nbfs(root);\r\nreturn 0;\r\n}

\n","title":"Breadth First Search Tree Traversal","authorDetails":{"name":"studymite","slug":"admin"},"description":"Breadth-First Search Traversal of a Tree using a queue data structute in iterative way.","slug":"breadth-first-search-tree-traversal/","articleId":"636a83417bb1ce5072622b05","bookmarkStatus":{"isBookmarked":false,"bookmarkId":null}}