We’ve discussed a simple iteration using two stacks after traversal in the previous article. In this article, the approach with only one stack is discussed.
The idea is to use the left pointer to move down to the leftmost node. When moving down, push the root and the right substack of the root. Once you reach the leftmost node, if you don’t have the right child node, you print it out. If it has the right child, change the root directory so that it can be processed before the right child is processed.
Here is the detailed algorithm.
1.1 Create an empty stack
2.1 Do following while root is not NULL
a) Push root's right child and then root to stack.
b) Set root as root's left child.
2.2 Pop an item from stack and set it as root.
a) If the popped item has a right child and the right child
is at top of stack, then remove the right child from stack, push the root back and set root as root's right child.
b) Else print root's data and set root as NULL.
2.3 Repeat steps 2.1 and 2.2 while stack is not empty.Let’s consider the tree below
Here are the steps to print a post-mortem traversal of the above tree using one stack.
1. Right child of 1 exists.
Push 3 to stack. Push 1 to stack. Move to left child.
Stack: 3, 1
2. Right child of 2 exists.
Push 5 to stack. Push 2 to stack. Move to left child.
Stack: 3, 1, 5, 2
3. Right child of 4 doesn't exist. '
Push 4 to stack. Move to left child.
Stack: 3, 1, 5, 2, 4
4. Current node is NULL.
Pop 4 from stack. Right child of 4 doesn't exist.
Print 4. Set current node to NULL.
Stack: 3, 1, 5, 2
5. Current node is NULL.
Pop 2 from stack. Since right child of 2 equals stack top element, pop 5 from stack. Now push 2 to stack.
Move current node to right child of 2 i.e. 5
Stack: 3, 1, 2
6. Right child of 5 doesn't exist. Push 5 to stack. Move to left child.
Stack: 3, 1, 2, 5
7. Current node is NULL. Pop 5 from stack. Right child of 5 doesn't exist.
Print 5. Set current node to NULL.
Stack: 3, 1, 2
8. Current node is NULL. Pop 2 from stack.
Right child of 2 is not equal to stack top element.
Print 2. Set current node to NULL.
Stack: 3, 1
9. Current node is NULL. Pop 1 from stack.
Since right child of 1 equals stack top element, pop 3 from stack.
Now push 1 to stack. Move current node to right child of 1 i.e. 3
Stack: 1
10. Repeat the same as above steps and Print 6, 7 and 3.
Pop 1 and Print 1.C
//C program for iterative postorder traversal using one stack
#include <stdio.h>
#include <stdlib.h>
//Maximum stack size
#define MAX_SIZE 100
//A tree node
struct Node
{
int data;
struct Node *left, *right;
};
//Stack type
struct Stack
{
int size;
int top;
struct Node* *array;
};
//A utility function to create a new tree node
struct Node* newNode( int data)
{
struct Node* node = ( struct Node*) malloc ( sizeof ( struct Node));
node->data = data;
node->left = node->right = NULL;
return node;
}
//A utility function to create a stack of given size
struct Stack* createStack( int size)
{
struct Stack* stack = ( struct Stack*) malloc ( sizeof ( struct Stack));
stack->size = size;
stack->top = -1;
stack->array = ( struct Node**) malloc (stack->size * sizeof ( struct Node*));
return stack;
}
//BASIC OPERATIONS OF STACK
int isFull( struct Stack* stack)
{ return stack->top - 1 == stack->size; }
int isEmpty( struct Stack* stack)
{ return stack->top == -1; }
void push( struct Stack* stack, struct Node* node)
{
if (isFull(stack))
return ;
stack->array[++stack->top] = node;
}
struct Node* pop( struct Stack* stack)
{
if (isEmpty(stack))
return NULL;
return stack->array[stack->top--];
}
struct Node* peek( struct Stack* stack)
{
if (isEmpty(stack))
return NULL;
return stack->array[stack->top];
}
//An iterative function to do postorder traversal of a given binary tree
void postOrderIterative( struct Node* root)
{
//Check for empty tree
if (root == NULL)
return ;
struct Stack* stack = createStack(MAX_SIZE);
do
{
//Move to leftmost node
while (root)
{
//Push root's right child and then root to stack.
if (root->right)
push(stack, root->right);
push(stack, root);
//Set root as root's left child
root = root->left;
}
//Pop an item from stack and set it as root
root = pop(stack);
//If the popped item has a right child and the right child is not
//processed yet, then make sure right child is processed before root
if (root->right && peek(stack) == root->right)
{
pop(stack); //remove right child from stack
push(stack, root); //push root back to stack
root = root->right; //change root so that the right
//child is processed next
}
else //Else print root's data and set root as NULL
{
printf ( "%d " , root->data);
root = NULL;
}
} while (!isEmpty(stack));
}
//Driver program to test above functions
int main()
{
//Let us construct the tree shown in above figure
struct Node* root = NULL;
root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->left = newNode(6);
root->right->right = newNode(7);
printf ( "Post order traversal of binary tree is :\n" );
printf ( "[" );
postOrderIterative(root);
printf ( "]" );
return 0;
}Java
//A java program for iterative postorder traversal using stack
import java.util.ArrayList;
import java.util.Stack;
//A binary tree node
class Node
{
int data;
Node left, right;
Node( int item)
{
data = item;
left = right;
}
}
class BinaryTree
{
Node root;
ArrayList<Integer> list = new ArrayList<Integer>();
//An iterative function to do postorder traversal
//of a given binary tree
ArrayList<Integer> postOrderIterative(Node node)
{
Stack<Node> S = new Stack<Node>();
//Check for empty tree
if (node == null )
return list;
S.push(node);
Node prev = null ;
while (!S.isEmpty())
{
Node current = S.peek();
/* go down the tree in search of a leaf an if so process it
and pop stack otherwise move down */
if (prev == null || prev.left == current ||
prev.right == current)
{
if (current.left != null )
S.push(current.left);
else if (current.right != null )
S.push(current.right);
else
{
S.pop();
list.add(current.data);
}
/* go up the tree from left node, if the child is right
push it onto stack otherwise process parent and pop
stack */
}
else if (current.left == prev)
{
if (current.right != null )
S.push(current.right);
else
{
S.pop();
list.add(current.data);
}
/* go up the tree from right node and after coming back
from right node process parent and pop stack */
}
else if (current.right == prev)
{
S.pop();
list.add(current.data);
}
prev = current;
}
return list;
}
//Driver program to test above functions
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
//Let us create trees shown in above diagram
tree.root = new Node( 1 );
tree.root.left = new Node( 2 );
tree.root.right = new Node( 3 );
tree.root.left.left = new Node( 4 );
tree.root.left.right = new Node( 5 );
tree.root.right.left = new Node( 6 );
tree.root.right.right = new Node( 7 );
ArrayList<Integer> mylist = tree.postOrderIterative(tree.root);
System.out.println( "Post order traversal of binary tree is :" );
System.out.println(mylist);
}
}
//This code has been contributed by Mayank JaiswalPython
# Python program for iterative postorder traversal
# using one stack
# Stores the answer
ans = []
# A Binary tree node
class Node:
# Constructor to create a new node
def __init__( self , data):
self .data = data
self .left = None
self .right = None
def peek(stack):
if len (stack)> 0 :
return stack[ - 1 ]
return None
# A iterative function to do postorder traversal of
# a given binary tree
def postOrderIterative(root):
# Check for empty tree
if root is None :
return
stack = []
while ( True ):
while (root):
# Push root's right child and then root to stack
if root.right is not None :
stack.append(root.right)
stack.append(root)
# Set root as root's left child
root = root.left
# Pop an item from stack and set it as root
root = stack.pop()
# If the popped item has a right child and the
# right child is not processed yet, then make sure
# right child is processed before root
if (root.right is not None and
peek(stack) = = root.right):
stack.pop() # Remove right child from stack
stack.append(root) # Push root back to stack
root = root.right # change root so that the
# righ childis processed next
# Else print root's data and set root as None
else :
ans.append(root.data)
root = None
if ( len (stack) <= 0 ):
break
# Driver pogram to test above function
root = Node( 1 )
root.left = Node( 2 )
root.right = Node( 3 )
root.left.left = Node( 4 )
root.left.right = Node( 5 )
root.right.left = Node( 6 )
root.right.right = Node( 7 )
print "Post Order traversal of binary tree is"
postOrderIterative(root)
print ans
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)The output is as follows:
Post Order traversal of binary tree is
[4, 5, 2, 6, 7, 3, 1]Method 2:
Move to the left twice, while pushing the root node directly twice. When ejected, if you find that stack top() is the same as root, then go to root-> right, otherwise print root.
//Simple Java program to print PostOrder Traversal(Iterative)
import java.util.Stack;
//A binary tree node
class Node
{
int data;
Node left, right;
Node( int item)
{
data = item;
left = right;
}
}
//create a postorder class
class PostOrder
{
Node root;
//An iterative function to do postorder traversal
//of a given binary tree
private void postOrderIterative(Node root) {
Stack<Node> stack = new Stack<>();
while ( true ) {
while (root != null ) {
stack.push(root);
stack.push(root);
root = root.left;
}
//Check for empty stack
if (stack.empty()) return ;
root = stack.pop();
if (!stack.empty() && stack.peek() == root) root = root.right;
else {
System.out.print(root.data + " " ); root = null ;
}
}
}
//Driver program to test above functions
public static void main(String args[])
{
PostOrder tree = new PostOrder();
//Let us create trees shown in above diagram
tree.root = new Node( 1 );
tree.root.left = new Node( 2 );
tree.root.right = new Node( 3 );
tree.root.left.left = new Node( 4 );
tree.root.left.right = new Node( 5 );
tree.root.right.left = new Node( 6 );
tree.root.right.right = new Node( 7 );
System.out.println( "Post order traversal of binary tree is :" );
tree.postOrderIterative(tree.root);
}
}The output is as follows:
Post Order traversal of binary tree is:
4, 5, 2, 6, 7, 3, 1This article was written by Aashish Barnwal. If you find anything incorrect, or would like to share more information about the above topics, please leave a comment.
