Inorder Non-threaded Binary Tree Traversal without Recursion or Stack

3.6

We have discussed Thread based Morris Traversal. Can we do inorder traversal without threads if we have parent pointers available to us?

Input: Root of Below Tree [Every node of 
       tree has parent pointer also]
        10
      /    \
     5     100
           /  \
          80  120 
Output: 5 10 80 100 120
The code should not extra space (No Recursion
and stack)

In inorder traversal, we follow “left root right”. We can move to children using left and right pointers. Once a node is visited, we need to move to parent also. For example, in the above tree, we need to move to 10 after printing 5. For this purpose, we use parent pointer. Below is algorithm.

1. Initialize current node as root
2. Initialize a flag: leftdone = false;
3. Do following while root is not NULL
     a) If leftdone is false, set current node as leftmost
        child of node. 
     b) Mark leftdone as true and print current node.
     c) If right child of current nodes exists, set current
        as right child and set leftdone as false.
     d) Else If parent exists, If current node is left child
        of its parent, set current node as parent. 
        If current node is right child, keep moving to ancestors
        using parent pointer while current node is right child
        of its parent.  
     e) Else break (We have reached back to root)

Illustration:

Let us consider below tree for illustration.
        10
      /    \
     5     100
           /  \
          80  120 

Initialize: Current node = 10, leftdone = false

Since leftdone is false, we move to 5 (3.a), print it
and set leftdone = true.

Now we move to parent of 5 (3.d). Node 10 is 
printed because leftdone is true.

We move to right of 10 and set leftdone as false (3.c)

Now current node is 100. Since leftdone is false, we move
to 80 (3.a) and set leftdone as true.  We print current 
node 80 and move back to parent 100 (3.d).  Since leftdone
is true, we print current node 100. 

Right of 100 exists, so we move to 120 (3.c).   We print
current node 120.

Since 120 is right child of its parent we keep moving to parent
while parent is right child of its parent.  We reach root. So
we break the loop and stop

Below is C++ implementation of above algorithm. Note that the implementation uses Binary Search Tree instead of Binary Tree. We can use the same function inorder() for Binary Tree also. The reason for using Binary Search Tree in below code is, it is easy to construct a Binary Search Tree with parent pointers and easy to test the outcome (In BST inorder traversal is always sorted).

C++

// C++ program to print inorder traversal of a Binary Search
// Tree (BST) without recursion and stack
#include <bits/stdc++.h>
using namespace std;

// BST Node
struct Node
{
    Node *left, *right, *parent;
    int key;
};

// A utility function to create a new BST node
Node *newNode(int item)
{
    Node *temp = new Node;
    temp->key = item;
    temp->parent = temp->left = temp->right = NULL;
    return temp;
}

/* A utility function to insert a new node with
   given key in BST */
Node *insert(Node *node, int key)
{
    /* If the tree is empty, return a new node */
    if (node == NULL) return newNode(key);

    /* Otherwise, recur down the tree */
    if (key < node->key)
    {
        node->left  = insert(node->left, key);
        node->left->parent = node;
    }
    else if (key > node->key)
    {
        node->right = insert(node->right, key);
        node->right->parent = node;
    }

    /* return the (unchanged) node pointer */
    return node;
}

// Function to print inorder traversal using parent
// pointer
void inorder(Node *root)
{
    bool leftdone = false;

    // Start traversal from root
    while (root)
    {
        // If left child is not traversed, find the
        // leftmost child
        if (!leftdone)
        {
            while (root->left)
                root = root->left;
        }

        // Print root's data
        printf("%d ", root->key);

        // Mark left as done
        leftdone = true;

        // If right child exists
        if (root->right)
        {
            leftdone = false;
            root = root->right;
        }

        // If right child doesn't exist, move to parent
        else if (root->parent)
        {
            // If this node is right child of its parent,
            // visit parent's parent first
            while (root->parent &&
                   root == root->parent->right)
                root = root->parent;
            if (!root->parent)
                break;
            root = root->parent;
        }
        else break;
    }
}

int main(void)
{
    Node * root = NULL;

    root = insert(root, 24);
    root = insert(root, 27);
    root = insert(root, 29);
    root = insert(root, 34);
    root = insert(root, 14);
    root = insert(root, 4);
    root = insert(root, 10);
    root = insert(root, 22);
    root = insert(root, 13);
    root = insert(root, 3);
    root = insert(root, 2);
    root = insert(root, 6);

    printf("Inorder traversal is \n");
    inorder(root);

    return 0;
}

Java

/* Java program to print inorder traversal of a Binary Search Tree
   without recursion and stack */
 
// BST node
class Node 
{
    int key;
    Node left, right, parent;
 
    public Node(int key) 
    {
        this.key = key;
        left = right = parent = null;
    }
}
 
class BinaryTree 
{
    Node root;
 
    /* A utility function to insert a new node with
       given key in BST */
    Node insert(Node node, int key) 
    {
        /* If the tree is empty, return a new node */
        if (node == null) 
            return new Node(key);
 
        /* Otherwise, recur down the tree */
        if (key < node.key) 
        {
            node.left = insert(node.left, key);
            node.left.parent = node;
        } 
        else if (key > node.key) 
        {
            node.right = insert(node.right, key);
            node.right.parent = node;
        }
         
        /* return the (unchanged) node pointer */
        return node;
    }
 
    // Function to print inorder traversal using parent
    // pointer
    void inorder(Node root) 
    {
        boolean leftdone = false;
 
        // Start traversal from root
        while (root != null) 
        {
            // If left child is not traversed, find the
            // leftmost child
            if (!leftdone) 
            {
                while (root.left != null) 
                {
                    root = root.left;
                }
            }
 
            // Print root's data
            System.out.print(root.key + " ");
 
            // Mark left as done
            leftdone = true;
 
            // If right child exists
            if (root.right != null) 
            {
                leftdone = false;
                root = root.right;
            } 
             
            // If right child doesn't exist, move to parent
            else if (root.parent != null) 
            {
                // If this node is right child of its parent,
                // visit parent's parent first
                while (root.parent != null
                        && root == root.parent.right) 
                    root = root.parent;
                 
                if (root.parent == null) 
                    break;
                root = root.parent;
            } 
            else
                break;
        }
    }
 
    public static void main(String[] args) 
    {
        BinaryTree tree = new BinaryTree();
        tree.root = tree.insert(tree.root, 24);
        tree.root = tree.insert(tree.root, 27);
        tree.root = tree.insert(tree.root, 29);
        tree.root = tree.insert(tree.root, 34);
        tree.root = tree.insert(tree.root, 14);
        tree.root = tree.insert(tree.root, 4);
        tree.root = tree.insert(tree.root, 10);
        tree.root = tree.insert(tree.root, 22);
        tree.root = tree.insert(tree.root, 13);
        tree.root = tree.insert(tree.root, 3);
        tree.root = tree.insert(tree.root, 2);
        tree.root = tree.insert(tree.root, 6);
 
        System.out.println("Inorder traversal is ");
        tree.inorder(tree.root);
    }
}
// This code has been contributed by Mayank Jaiswal(mayank_24)

Output:

Inorder traversal is 
2 3 4 6 10 13 14 22 24 27 29 34

This article is contributed by Rishi Chhibber. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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