Reputation: 392
I am quite confused how to maintain the top of tree while traversing through BFS. I am accepting input like this :
5 // total number of nodes in tree (root node will always be 1)
// this number also tells the number of lines that'll be coming up
2 3 // left and right child of root, tree be like : 1
/ \
2 3
4 5 // left and right child of 2 , tree 1
/ \
2 3
/ \
4 5
-1 -1 // left and right child of 3 , that's null
-1 -1 // left and right child of 4
-1 -1 // left and right child of 5
This continues in above fashion, user inputs the left and right child in BFS. But I am unable to comprehend how do I achieve this.
What my take was :
LinkedList<Node> list = new LinkedList<Node>
list.add(root); //initially push root
while(list.peek()){ //while there is node left in linked list
curr = list.poll();
curr.left = new Node(x) // x = whatever is the user input is (using nextInt)
// and if it's -1, i'll make left child null
curr.right = new Node(x) // same goes for this one
ll.add(curr);
}
In the end, I need the reference to the root node, I don't know how do I get it? And also, is there any better method how I could achieve this task
Upvotes: 0
Views: 1186
Reputation: 203
I hope below code might help you.
"readTree" function is used to reading tree.
public static Node readTree(Scanner in) {
Queue<Node> queue = new LinkedList<Node>();
Node root = new Node(1);
queue.add(root);
while (!queue.isEmpty()) {
Node node = queue.poll();
int left = in.nextInt();
int right = in.nextInt();
if (-1 != left) {
node.left = new Node(left);
queue.add(node.left);
}
if (-1 != right) {
node.right = new Node(right);
queue.add(node.right);
}
}
return root;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
in.nextInt(); // number of nodes is not used.
Node result = readTree(in);
}
Upvotes: 1