Problem Statement:
Print Level order traversal of Binary Tree (Using Queue)
Required Output: 1 2 3 4 5 6 7 8 9 10 11 12
Solution :
Level order traversal of a binary tree also means BFS (Breadth First Search) , in which each children of a node is processed , then the the subsequent nodes unlike DFS(Depth First Search).
Please refer Source Code here
Please post comments if any suggestion or discrepancy.
Happy Coding !! :)
Print Level order traversal of Binary Tree (Using Queue)
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| Fig Level order traversal of Binary Tree |
Required Output: 1 2 3 4 5 6 7 8 9 10 11 12
Solution :
Level order traversal of a binary tree also means BFS (Breadth First Search) , in which each children of a node is processed , then the the subsequent nodes unlike DFS(Depth First Search).
Please refer Source Code here
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 | public void printLevelOrder(Node root){ if(root==null) return ; LinkedList<Node> queue = new LinkedList<Node>(); queue.addLast(root); queue.addLast(null); while(!queue.isEmpty()){ Node poped = queue.removeFirst(); //dequeue if (poped == null) { if(queue.isEmpty()) // if last node , terminate continue; queue.addLast(null); System.out.println(); } else { System.out.print(poped.data + "\t"); if (poped.left != null) queue.addLast(poped.left); if (poped.right != null) queue.addLast(poped.right); } } } |
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