Print a binary tree level by level

Question

THY · Accepted Answer

public Void BFS()    
{      
 Queue q = new Queue();
 q.Enqueue(root);//You don't need to write the root here, it will be written in the loop
 while (q.count > 0)
 {
    Node n = q.DeQueue();
    Console.Writeln(n.Value); //Only write the value when you dequeue it
    if (n.left !=null)
    {
        q.EnQueue(n.left);//enqueue the left child
    }
    if (n.right !=null)
    {
       q.EnQueue(n.right);//enque the right child
    }
 }
}

kilaka · Answer

See oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/

kilaka · Answer

Oops. I meant this: oj.leetcode.com/problems/binary-tree-level-order-traversal/

someone · Answer

Improving THY answer:


public Void BFS(Node root)
{
if(root == null)
    return;
 Queue q = new Queue();
 q.Enqueue(root);//You don't need to write the root here, it will be written in the loop
 while (!q.empty())
 {
    Node current = q.DeQueue();
    Console.Writeln(n.Value); //Only write the value when you dequeue it
    if (n.left !=null){  q.EnQueue(n.left);} //enqueue the left child
    
    if (n.right !=null) { q.EnQueue(n.right); } //enque the right child

    q.pop(); // popping since we used the dequed node
 }
}

kilaka · Answer

Darn, can't edit comments. The oj.leetcode link links to something similar but not exactly.

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