Inorder Non-threaded Binary Tree Traversal without Recursion or Stack
Last Updated :
17 Jul, 2022
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 the 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 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)
# Python3 program to print inorder traversal of a
# Binary Search Tree (BST) without recursion and stack
# A utility function to create a new BST node
class newNode:
def __init__(self, item):
self.key = item
self.parent = self.left = self.right = None
# A utility function to insert a new
# node with given key in BST
def insert(node, key):
# If the tree is empty, return a new node
if node == None:
return newNode(key)
# Otherwise, recur down the tree
if key < node.key:
node.left = insert(node.left, key)
node.left.parent = node
elif 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
def inorder(root):
leftdone = False
# Start traversal from root
while root:
# If left child is not traversed,
# find the leftmost child
if leftdone == False:
while root.left:
root = root.left
# Print root's data
print(root.key, end = " ")
# 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
elif root.parent:
# If this node is right child of its
# parent, visit parent's parent first
while root.parent and root == root.parent.right:
root = root.parent
if root.parent == None:
break
root = root.parent
else:
break
# Driver Code
if __name__ == '__main__':
root = None
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)
print("Inorder traversal is ")
inorder(root)
# This code is contributed by PranchalK
// C# program to print inorder traversal
// of a Binary Search Tree without
// recursion and stack
using System;
// BST node
class Node
{
public int key;
public 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
Console.Write(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;
}
}
// Driver Code
static public 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);
Console.WriteLine("Inorder traversal is ");
tree.inorder(tree.root);
}
}
// This code is contributed by Arnab Kundu
<script>
/* javascript program to print inorder traversal of a Binary Search Tree
without recursion and stack */
// BST node
class Node {
constructor(key) {
this.key = key;
this.left = this.right = this.parent = null;
}
}
var root = null;
/*
* A utility function to insert a new node with given key in BST
*/
function insert(node , 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
function inorder(root) {
var 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
document.write(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;
}
}
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);
document.write("Inorder traversal is ");
inorder(root);
// This code contributed by aashish1995
</script>
Output:
Inorder traversal is
2 3 4 6 10 13 14 22 24 27 29 34
Time complexity: O(n) where n is no of nodes
Auxiliary Space: O(n)
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