How to construct a binary tree using a level order traversal sequence

Question

How to construct a binary tree using a level order traversal sequence, for example from sequence {1,2,3,#,#,4,#,#,5}, we can construct a binary tree like this:

Answer 1

Assume using array int[]data with 0-based index, we have a simple function to get children:

• Left child

  int getLeftChild(int index){
     if(index*2 + 1 >= data.length)
        return -1;// -1 Means out of bound
     return data[(index*2) + 1];
  }
  

• Right child

  int getRightChild(int index){
     if(index*2 + 2 >= data.length)
        return -1;// -1 Means out of bound           
     return data[(index*2) + 2];
  }
  

Edit: Ok, so by maintaining a queue, we can build this binary tree.

We use a queue to maintain those nodes that are not yet processed.

Using a variable count to keep track of the number of children added for the current node.

First, create a root node, assign it as the current node. So starting from index 1 (index 0 is the root), as the count is 0, we add this node as left child of the current node. Increase count. If this node is not '#', add it to the queue.

Moving to the next index, the count is 1, so we add this as right child of current node, reset count to 0 and update current node (by assigning the current node as the first element in the queue). If this node is not '#', add it to the queue.

     int count = 0;
     Queue q = new Queue();
     q.add(new Node(data[0]);
     Node cur = null;
     for(int i = 1; i < data.length; i++){
        Node node = new Node(data[i]);
        if(count == 0){
           cur = q.dequeue();           
        }
        if(count==0){
          count++;
          cur.leftChild = node;
        }else {
          count = 0;
          cur.rightChild = node;
        }
        if(data[i] != '#'){
          q.enqueue(node);
        }
     }    



    class Node{
       int data;
       Node leftChild, rightChild;
    }