Given a binary tree, print out the nodes in order without using recursion. - Java Data Structure

Java examples for Data Structure:Tree

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Description

Given a binary tree, print out the nodes in order without using recursion.

Demo Code

import java.util.Stack;

public class InorderTraversal {
    public static class Node {
        Node left;/*from   w  w w  .  j  av a2  s.c  o  m*/
        Node right;
        int value;

        public Node(int value) {
            this.value = value;
        }
    }

    // Traverse tree iteratively. We do this by replicating the same process
    // done recursively using an explicit stack
    public static void traverse(Node n) {
        Stack<Node> s = new Stack<Node>();
        // Add the leftmost branch to the stack
        addLeftToStack(s, n);

        // While there are elements in the stack, pop and move the minimum
        // possible distance to the right
        while (!s.isEmpty()) {
            Node curr = s.pop();
            System.out.println(curr.value);

            // Add the right child and any of its left children to the stack
            addLeftToStack(s, curr.right);
        }
    }

    // As long as the current node has a left pointer, add it to the stack and 
    // continue
    private static void addLeftToStack(Stack<Node> s, Node n) {
        while (n != null) {
            s.push(n);
            n = n.left;
        }
    }

    /*
     * Sample test case
     * 
     *        8
     *     /     \
     *    4       12
     *   / \     /  \
     *  1   6   9   15
     *     /     \
     *    5      10
     * 
     * Should print:
     * 1
     * 4
     * 5
     * 6
     * 8
     * 9
     * 10
     * 12
     * 15
     */

    public static void main(String[] args) {
        Node n = new Node(8);
        n.left = new Node(4);
        n.left.left = new Node(1);
        n.left.right = new Node(6);
        n.left.right.left = new Node(5);
        n.right = new Node(12);
        n.right.left = new Node(9);
        n.right.left.right = new Node(10);
        n.right.right = new Node(15);
        traverse(n);
    }
}

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