Level Order Traversal w/ Binary Trees using Java and Recursion

问题

I am having problems with the level order traversal of my binary tree whilst using recursion. I am inputing the following values: 50,60,70,30,20,10 Here is the code I am using:

    public void levelOrder(Node localRoot){
    if(localRoot != null){
        if(localRoot.leftChild != null && localRoot.rightChild != null){
            System.out.print(localRoot.iData + " ");
            System.out.print(localRoot.leftChild.iData + " ");
            System.out.print(localRoot.rightChild.iData + " ");
            levelOrder(localRoot.leftChild);
            levelOrder(localRoot.rightChild);
        }
        else if(localRoot.rightChild == null && localRoot.leftChild == null){
            ;
        }
        else if(localRoot.rightChild == null){
            System.out.print(localRoot.leftChild.iData + " ");
            //levelOrder(localRoot.leftChild);
        }
        else{
            System.out.print(localRoot.rightChild.iData + " ");
            //levelOrder(localRoot.rightChild);
        }
    }
}

Is recursion possible without using a stack? Because currently this function is taking me all the way to the left and then it goes right. What could I be doing differently?

My output for this code is: 50, 30, 60, 20, 70 and it does not print 10.

回答1:

Interestingly this is a fairly common question (google "recursive bread first traversal" and there are several links to stackoverflow of similar answers )

by far the best one is here

Performing Breadth First Search recursively

and i agree with the author of the top answer, there is really no point in turning an iterative algorithm (breadth first traversal) into a recursive solution. As mentioned yes its easy to turn iteration into tail recursion, but to what end? and you would still need a queue.

回答2:

public void level() {

 printLevel(root);

}

public void printLevel(Node current)
{
    if(current != null){
        if(current.leftChild != null && current.rightChild != null){
            System.out.print(current.iData + " ");
            System.out.print(current.leftChild.iData + " ");
            System.out.print(current.rightChild.iData + " ");
            printLevel(current.leftChild);
            printLevel(current.rightChild);
        }
        else if(current.rightChild == null && current.leftChild == null){
        ;
        }
        else if(current.rightChild == null){
            System.out.print(current.leftChild.iData + " ");
            //levelOrder(current.leftChild);
        }
        else{
            System.out.print(current.rightChild.iData + " ");
            //levelOrder(current.rightChild);
        }
    }
}

回答3:

You can do the trick with a Queue:

private Queue<Key> getLevelOrderKeys(Node root){

    if (root == null) {
        return null;
    }

    Queue<Key> keyQueue = new ArrayDeque<Key>(); //return value. Queue with the level order keys
    Queue<Node> nodeQueue = new ArrayDeque<Node>();

    nodeQueue.add(root);

    //while there is at least one discovered node
    while(!nodeQueue.isEmpty()) {

        Node current = nodeQueue.poll();
        keyQueue.add(current.key);

        if (current.left != null){
            nodeQueue.add(current.left);
        }

        if (current.right != null){
            nodeQueue.add(current.right);
        }

    }

    return keyQueue;

}

来源：https://stackoverflow.com/questions/19964048/level-order-traversal-w-binary-trees-using-java-and-recursion

标签

java

recursion

binary-tree