What is post-order traversal?

Tree traversal refers to visiting all the nodes of a tree exactly once. Visiting means doing something to the node. It can be as basic as printing the node.

Post-order traversal is one of the multiple methods to traverse a tree. It is mainly used for tree deletion.

Basic Algorithm

The following algorithm is specific to a binary tree but can be generalized​ to an n-ary tree (a tree where each node can have at most​ n children nodes).

1. Recursively traverse the left subtree.
2. Recursively traverse the right subtree.
3. Visit the root.

Unlike linear data structures (Array, Linked List, Queues, Stacks, etc) which have only one logical way to traverse them, trees can be traversed in different ways. 

A Tree Data Structure can be traversed in following ways:

1. Depth First Search or DFS
   1. Inorder Traversal
   2. Preorder Traversal
   3. Postorder Traversal
2. Level Order Traversal or Breadth First Search or BFS
3. Boundary Traversal
4. Diagonal Traversal

Algorithm Inorder(tree)

1. Traverse the left subtree, i.e., call Inorder(left->subtree)
2. Visit the root.
3. Traverse the right subtree, i.e., call Inorder(right->subtree)

Uses of Inorder Traversal:

In the case of binary search trees (BST), Inorder traversal gives nodes in non-decreasing order. To get nodes of BST in non-increasing order, a variation of Inorder traversal where Inorder traversal is reversed can be used.

Code implementation of Inorder traversal.

C

#include <stdio.h>

#include <stdlib.h>

 

struct node {

    int data;

    struct node* left;

    struct node* right;

};

 

struct node* newNode(int data)

{

    struct node* node

        = (struct node*)malloc(sizeof(struct node));

    node->data = data;

    node->left = NULL;

    node->right = NULL;

 

    return (node);

}

 

void printInorder(struct node* node)

{

    if (node == NULL)

        return;

 

    

    printInorder(node->left);

 

    

    printf("%d ", node->data);

 

    

    printInorder(node->right);

}

 

int main()

{

    struct node* root = newNode(1);

    root->left = newNode(2);

    root->right = newNode(3);

    root->left->left = newNode(4);

    root->left->right = newNode(5);

 

    

    printf("Inorder traversal of binary tree is \n");

    printInorder(root);

 

    getchar();

    return 0;

}

C++

#include <bits/stdc++.h>

using namespace std;

 

struct Node {

    int data;

    struct Node *left, *right;

};

 

Node* newNode(int data)

{

    Node* temp = new Node;

    temp->data = data;

    temp->left = temp->right = NULL;

    return temp;

}

 

void printInorder(struct Node* node)

{

    if (node == NULL)

        return;

 

    

    printInorder(node->left);

 

    

    cout << node->data << " ";

 

    

    printInorder(node->right);

}

 

int main()

{

    struct Node* root = newNode(1);

    root->left = newNode(2);

    root->right = newNode(3);

    root->left->left = newNode(4);

    root->left->right = newNode(5);

 

    

    cout << "Inorder traversal of binary tree is \n";

    printInorder(root);

 

    return 0;

}

Java

 

class Node {

    int key;

    Node left, right;

 

    public Node(int item)

    {

        key = item;

        left = right = null;

    }

}

 

class BinaryTree {

 

    

    Node root;

 

    BinaryTree() { root = null; }

 

    

    void printInorder(Node node)

    {

        if (node == null)

            return;

 

        

        printInorder(node.left);

 

        

        System.out.print(node.key + " ");

 

        

        printInorder(node.right);

    }

 

    

    public static void main(String[] args)

    {

        BinaryTree tree = new BinaryTree();

        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);

 

        

        System.out.println(

            "Inorder traversal of binary tree is ");

        tree.printInorder(tree.root);

    }

}

Python3

 

 

class Node:

    def __init__(self, key):

        self.left = None

        self.right = None

        self.val = key

 

 

def printInorder(root):

 

    if root:

 

        

        printInorder(root.left)

 

        

        print(root.val, end=" "),

 

        

        printInorder(root.right)

 

 

if __name__ == "__main__":

    root = Node(1)

    root.left = Node(2)

    root.right = Node(3)

    root.left.left = Node(4)

    root.left.right = Node(5)

 

    

    print("Inorder traversal of binary tree is")

    printInorder(root)

C#

using System;

 

class Node {

    public int key;

    public Node left, right;

 

    public Node(int item)

    {

        key = item;

        left = right = null;

    }

}

 

class BinaryTree {

 

    

    Node root;

 

    BinaryTree() { root = null; }

 

    

    

    void printInorder(Node node)

    {

        if (node == null)

            return;

 

        

        printInorder(node.left);

 

        

        Console.Write(node.key + " ");

 

        

        printInorder(node.right);

    }

 

    

    void printInorder() { printInorder(root); }

 

    

    static public void Main(String[] args)

    {

        BinaryTree tree = new BinaryTree();

        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);

 

        

        Console.WriteLine("Inorder traversal "

                          + "of binary tree is ");

        tree.printInorder();

    }

}

 

Javascript

 

class Node {

    constructor(val) {

        this.key = val;

        this.left = null;

        this.right = null;

    }

}

 

var root = null;

     

function printInorder(node) {

    if (node == null)

        return;

 

    

    printInorder(node.left);

 

    

    console.log(node.key + " ");

 

    

    printInorder(node.right);

}

 

     

    root = new Node(1);

    root.left = new Node(2);

    root.right = new Node(3);

    root.left.left = new Node(4);

    root.left.right = new Node(5);

 

    console.log("Inorder traversal of binary tree is ");

    printInorder(root);

 

Output

Inorder traversal of binary tree is 
4 2 5 1 3 

Time Complexity: O(N)
Auxiliary Space: If we don’t consider the size of the stack for function calls then O(1) otherwise O(h) where h is the height of the tree. 

Algorithm Preorder(tree)

1. Visit the root.
2. Traverse the left subtree, i.e., call Preorder(left->subtree)
3. Traverse the right subtree, i.e., call Preorder(right->subtree) 

Uses of Preorder:

Preorder traversal is used to create a copy of the tree. Preorder traversal is also used to get prefix expressions on an expression tree.

Code implementation of Preorder traversal:

C

#include <stdio.h>

#include <stdlib.h>

 

struct node {

    int data;

    struct node* left;

    struct node* right;

};

 

struct node* newNode(int data)

{

    struct node* node

        = (struct node*)malloc(sizeof(struct node));

    node->data = data;

    node->left = NULL;

    node->right = NULL;

 

    return (node);

}

 

void printPreorder(struct node* node)

{

    if (node == NULL)

        return;

 

    

    printf("%d ", node->data);

 

    

    printPreorder(node->left);

 

    

    printPreorder(node->right);

}

 

int main()

{

    struct node* root = newNode(1);

    root->left = newNode(2);

    root->right = newNode(3);

    root->left->left = newNode(4);

    root->left->right = newNode(5);

 

    

    printf("Preorder traversal of binary tree is \n");

    printPreorder(root);

 

    getchar();

    return 0;

}

C++

#include <bits/stdc++.h>

using namespace std;

 

struct Node {

    int data;

    struct Node *left, *right;

};

 

Node* newNode(int data)

{

    Node* temp = new Node;

    temp->data = data;

    temp->left = temp->right = NULL;

    return temp;

}

 

void