We have discussed a simple iterative postorder traversal using two stacks in the previous post. In this post, an approach with only one stack is discussed.
The idea is to move down to leftmost node using left pointer. While moving down, push root and root’s right child to stack. Once we reach leftmost node, print it if it doesn’t have a right child. If it has a right child, then change root so that the right child is processed before.
Following is 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 us consider the following tree
Following are the steps to print postorder 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; } |
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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 Jaiswal |
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Python
# 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) |
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Output:
Post Order traversal of binary tree is [4, 5, 2, 6, 7, 3, 1]
Method 2:
Push directly root node two times while traversing to the left. While poping if you find stack top() is same as root then go for root->right else 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); } } |
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Output:
Post Order traversal of binary tree is: 4, 5, 2, 6, 7, 3, 1
This article is compiled by Aashish Barnwal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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Improved By : ashishfk
