Traversing through all nodes of a binary tree in Java

Question

Let's say I have a simple binary tree node class, like so:

public class BinaryTreeNode {
    public String identifier = "";
    public BinaryTreeNode parent = null;
    public BinaryTreeNode left = null;
    public BinaryTreeNode right = null;

    public BinaryTreeNode(BinaryTreeNode parent, String identifier)
    {
        this.parent = parent; //passing null makes this the root node
        this.identifier = identifier;
    }

    public boolean IsRoot() {
        return parent == null;
    }
}

How would I add a method which is able to recursively traverse through any size tree, visiting each and every existing node from left to right, without revisiting a node that has already been traversed?

Would this work?:

public void traverseFrom(BinaryTreeNode rootNode)
{
    /* insert code dealing with this node here */

    if(rootNode.left != null)
        rootNode.left.traverseFrom(rootNode.left);

    if(rootNode.right != null)
        rootNode.traverseFrom(rootNode.right);
}

Answer 1

There are 3 types of Binary tree traversal that you can achieve :

example:

consider this following Binary tree :

Pre-order traversal sequence: F, B, A, D, C, E, G, I, H (root, left, right)
In-order traversal sequence: A, B, C, D, E, F, G, H ,I (left, root, right)
Post-order traversal sequence: A, C, E, D, B, H, I, G, F (left, right, root)

code example:

left to right traversal of the Binary tree, nay In order Traversal of binary tree :

public void traverse (Node root){ // Each child of a tree is a root of its subtree.
    if (root.left != null){
        traverse (root.left);
    }
    System.out.println(root.data);
    if (root.right != null){
        traverse (root.right);
    }
}

Answer 2

codeMan is right. The traversal will visit every node on the left. Once it reaches the last node on the left, it begins working its way back along the right-side nodes. This is a depth-first search (DFS) traversal. As such, each node is visited only once, and the algorithm runs in O(n) time. Happy coding.