Skip to content
Related Articles

Related Articles

Tree Traversals (Inorder, Preorder and Postorder)

View Discussion
Improve Article
Save Article
  • Difficulty Level : Easy
  • Last Updated : 21 Sep, 2022
View Discussion
Improve Article
Save Article

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. Following are the generally used methods for traversing trees:

Example:

Inorder Traversal (Practice): 

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. 
Example: In order traversal for the above-given figure is 4 2 5 1 3.

C++




// C++ program for different tree traversals
#include <bits/stdc++.h>
using namespace std;
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct Node {
    int data;
    struct Node *left, *right;
};
 
// Utility function to create a new tree node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
/* Given a binary tree, print its nodes in inorder*/
void printInorder(struct Node* node)
{
    if (node == NULL)
        return;
 
    /* first recur on left child */
    printInorder(node->left);
 
    /* then print the data of node */
    cout << node->data << " ";
 
    /* now recur on right child */
    printInorder(node->right);
}
 
/* Driver code*/
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);
 
 
      // Function call
    cout << "\nInorder traversal of binary tree is \n";
    printInorder(root);
 
 
    return 0;
}


C




// C program for different tree traversals
#include <stdio.h>
#include <stdlib.h>
 
/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct node {
    int data;
    struct node* left;
    struct node* right;
};
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
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);
}
 
/* Given a binary tree, print its nodes in inorder*/
void printInorder(struct node* node)
{
    if (node == NULL)
        return;
 
    /* first recur on left child */
    printInorder(node->left);
 
    /* then print the data of node */
    printf("%d ", node->data);
 
    /* now recur on right child */
    printInorder(node->right);
}
 
 
/* Driver code*/
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);
 
      // Function call
    printf("\nInorder traversal of binary tree is \n");
    printInorder(root);
 
    getchar();
    return 0;
}


Java




// Java program for different tree traversals
 
/* Class containing left and right child of current
   node and key value*/
class Node {
    int key;
    Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
class BinaryTree {
    // Root of Binary Tree
    Node root;
 
    BinaryTree() { root = null; }
 
    /* Given a binary tree, print its nodes in inorder*/
    void printInorder(Node node)
    {
        if (node == null)
            return;
 
        /* first recur on left child */
        printInorder(node.left);
 
        /* then print the data of node */
        System.out.print(node.key + " ");
 
        /* now recur on right child */
        printInorder(node.right);
    }
 
    // Wrappers over above recursive functions
    void printInorder() { printInorder(root); }
 
    // Driver code
    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);
 
          // Function call
        System.out.println(
            "\nInorder traversal of binary tree is ");
        tree.printInorder();
    }
}


Python




# Python3 program to for tree traversals
 
# A class that represents an individual node in a
# Binary Tree
 
 
class Node:
    def __init__(self, key):
        self.left = None
        self.right = None
        self.val = key
 
 
# A function to do inorder tree traversal
def printInorder(root):
 
    if root:
 
        # First recur on left child
        printInorder(root.left)
 
        # then print the data of node
        print(root.val),
 
        # now recur on right child
        printInorder(root.right)
 
 
# Driver code
if __name__ == "__main__":
  root = Node(1)
  root.left = Node(2)
  root.right = Node(3)
  root.left.left = Node(4)
  root.left.right = Node(5)
 
  # Function call
  print "\nInorder traversal of binary tree is"
  printInorder(root)


C#




// C# program for different
// tree traversals
using System;
 
/* Class containing left and
right child of current
node and key value*/
class Node {
    public int key;
    public Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
class BinaryTree {
    // Root of Binary Tree
    Node root;
 
    BinaryTree() { root = null; }
 
    /* Given a binary tree, print
       its nodes in inorder*/
    void printInorder(Node node)
    {
        if (node == null)
            return;
 
        /* first recur on left child */
        printInorder(node.left);
 
        /* then print the data of node */
        Console.Write(node.key + " ");
 
        /* now recur on right child */
        printInorder(node.right);
    }
 
    // Wrappers over above recursive functions
    void printInorder() { printInorder(root); }
 
    // Driver Code
    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);
 
          // Function call
        Console.WriteLine("\nInorder traversal "
                          + "of binary tree is ");
        tree.printInorder();
    }
}
 
// This code is contributed by Arnab Kundu


Javascript




<script>
// javascript program for different tree traversals
 
/* Class containing left and right child of current
   node and key value*/
class Node {
    constructor(val) {
        this.key = val;
        this.left = null;
        this.right = null;
    }
}
 
    // Root of Binary Tree
    var root = null;
 
     
    /*
     * Given a binary tree, print its nodes according to the "bottom-up" postorder
     * traversal.
     */
    function printPostorder(node) {
        if (node == null)
            return;
 
        // first recur on left subtree
        printPostorder(node.left);
 
        // then recur on right subtree
        printPostorder(node.right);
 
        // now deal with the node
        document.write(node.key + " ");
    }
 
    /* Given a binary tree, print its nodes in inorder */
    function printInorder(node) {
        if (node == null)
            return;
 
        /* first recur on left child */
        printInorder(node.left);
 
        /* then print the data of node */
        document.write(node.key + " ");
 
        /* now recur on right child */
        printInorder(node.right);
    }
 
    /* Given a binary tree, print its nodes in preorder */
    function printPreorder(node) {
        if (node == null)
            return;
 
        /* first print data of node */
        document.write(node.key + " ");
 
        /* then recur on left subtree */
        printPreorder(node.left);
 
        /* now recur on right subtree */
        printPreorder(node.right);
        
    }
 
 
 
    // Driver method
     
     
        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);
 
        document.write("Preorder traversal of binary tree is <br/>");
        printPreorder(root);
 
        document.write("<br/>Inorder traversal of binary tree is <br/>");
        printInorder(root);
 
        document.write("<br/>Postorder traversal of binary tree is <br/>");
        printPostorder(root);
 
// This code is contributed by aashish1995
</script>


Preorder Traversal (Practice): 

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.
Example: Preorder traversal for the above-given figure is 1 2 4 5 3.

C++




// C++ program for different tree traversals
#include <bits/stdc++.h>
using namespace std;
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct Node {
    int data;
    struct Node *left, *right;
};
 
// Utility function to create a new tree node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
/* Given a binary tree, print its nodes in preorder*/
void printPreorder(struct Node* node)
{
    if (node == NULL)
        return;
 
    /* first print data of node */
    cout << node->data << " ";
 
    /* then recur on left subtree */
    printPreorder(node->left);
 
    /* now recur on right subtree */
    printPreorder(node->right);
}
 
/* Driver code*/
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);
 
      // Function call
    cout << "\nPreorder traversal of binary tree is \n";
    printPreorder(root);
 
    return 0;
}


C




// C program for different tree traversals
#include <stdio.h>
#include <stdlib.h>
 
/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct node {
    int data;
    struct node* left;
    struct node* right;
};
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
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);
}
 
/* Given a binary tree, print its nodes in preorder*/
void printPreorder(struct node* node)
{
    if (node == NULL)
        return;
 
    /* first print data of node */
    printf("%d ", node->data);
 
    /* then recur on left subtree */
    printPreorder(node->left);
 
    /* now recur on right subtree */
    printPreorder(node->right);
}
 
/* Driver code*/
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);
 
      // Function call
    printf("\nPreorder traversal of binary tree is \n");
    printPreorder(root);
 
    getchar();
    return 0;
}


Java




// Java program for different tree traversals
 
/* Class containing left and right child of current
   node and key value*/
class Node {
    int key;
    Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
class BinaryTree {
    // Root of Binary Tree
    Node root;
 
    BinaryTree() { root = null; }
   
    /* Given a binary tree, print its nodes in preorder*/
    void printPreorder(Node node)
    {
        if (node == null)
            return;
 
        /* first print data of node */
        System.out.print(node.key + " ");
 
        /* then recur on left subtree */
        printPreorder(node.left);
 
        /* now recur on right subtree */
        printPreorder(node.right);
    }
 
    // Wrappers over above recursive functions
    void printPreorder() { printPreorder(root); }
 
    // Driver code
    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);
 
          // Function call
        System.out.println(
            "Preorder traversal of binary tree is ");
        tree.printPreorder();
    }
}


Python




# Python3 program to for tree traversals
 
# A class that represents an individual node in a
# Binary Tree
 
 
class Node:
    def __init__(self, key):
        self.left = None
        self.right = None
        self.val = key
 
 
# A function to do preorder tree traversal
def printPreorder(root):
 
    if root:
 
        # First print the data of node
        print(root.val),
 
        # Then recur on left child
        printPreorder(root.left)
 
        # Finally recur on right child
        printPreorder(root.right)
 
 
# Driver code
if __name__ == "__main__":
  root = Node(1)
  root.left = Node(2)
  root.right = Node(3)
  root.left.left = Node(4)
  root.left.right = Node(5)
 
  # Function call
  print "Preorder traversal of binary tree is"
  printPreorder(root)


C#




// C# program for different
// tree traversals
using System;
 
/* Class containing left and
right child of current
node and key value*/
class Node {
    public int key;
    public Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
class BinaryTree {
    // Root of Binary Tree
    Node root;
 
    BinaryTree() { root = null; }
 
    /* Given a binary tree, print
       its nodes in preorder*/
    void printPreorder(Node node)
    {
        if (node == null)
            return;
 
        /* first print data of node */
        Console.Write(node.key + " ");
 
        /* then recur on left subtree */
        printPreorder(node.left);
 
        /* now recur on right subtree */
        printPreorder(node.right);
    }
 
    // Wrappers over above recursive functions
    void printPreorder() { printPreorder(root); }
 
    // Driver Code
    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);
 
          // Function call
        Console.WriteLine("Preorder traversal "
                          + "of binary tree is ");
        tree.printPreorder();
    }
}
 
// This code is contributed by Arnab Kundu


Javascript




<script>
// javascript program for different tree traversals
 
/* Class containing left and right child of current
   node and key value*/
class Node {
    constructor(val) {
        this.key = val;
        this.left = null;
        this.right = null;
    }
}
 
    // Root of Binary Tree
    var root = null;
 
     
    /*
     * Given a binary tree, print its nodes according to the "bottom-up" postorder
     * traversal.
     */
    function printPostorder(node) {
        if (node == null)
            return;
 
        // first recur on left subtree
        printPostorder(node.left);
 
        // then recur on right subtree
        printPostorder(node.right);
 
        // now deal with the node
        document.write(node.key + " ");
    }
 
    /* Given a binary tree, print its nodes in inorder */
    function printInorder(node) {
        if (node == null)
            return;
 
        /* first recur on left child */
        printInorder(node.left);
 
        /* then print the data of node */
        document.write(node.key + " ");
 
        /* now recur on right child */
        printInorder(node.right);
    }
 
    /* Given a binary tree, print its nodes in preorder */
    function printPreorder(node) {
        if (node == null)
            return;
 
        /* first print data of node */
        document.write(node.key + " ");
 
        /* then recur on left subtree */
        printPreorder(node.left);
 
        /* now recur on right subtree */
        printPreorder(node.right);
        
    }
 
 
 
    // Driver method
     
     
        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);
 
        document.write("Preorder traversal of binary tree is <br/>");
        printPreorder(root);
 
        document.write("<br/>Inorder traversal of binary tree is <br/>");
        printInorder(root);
 
        document.write("<br/>Postorder traversal of binary tree is <br/>");
        printPostorder(root);
 
// This code is contributed by aashish1995
</script>


Postorder Traversal (Practice): 

Algorithm Postorder(tree)

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

Uses of Postorder:

Postorder traversal is used to delete the tree. Please see the question for the deletion of a tree for details. Postorder traversal is also useful to get the postfix expression of an expression tree
Example: Postorder traversal for the above-given figure is 4 5 2 3 1

Below is the implementation of the above traversal methods:

C++




// C++ program for different tree traversals
#include <bits/stdc++.h>
using namespace std;
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct Node {
    int data;
    struct Node *left, *right;
};
 
// Utility function to create a new tree node
Node* newNode(int data)
{
    Node* temp = new Node;
    temp->data = data;
    temp->left = temp->right = NULL;
    return temp;
}
 
/* Given a binary tree, print its nodes according to the
"bottom-up" postorder traversal. */
void printPostorder(struct Node* node)
{
    if (node == NULL)
        return;
 
    // first recur on left subtree
    printPostorder(node->left);
 
    // then recur on right subtree
    printPostorder(node->right);
 
    // now deal with the node
    cout << node->data << " ";
}
 
/* Driver code*/
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);
 
      // Function call
    cout << "\nPostorder traversal of binary tree is \n";
    printPostorder(root);
 
    return 0;
}


C




// C program for different tree traversals
#include <stdio.h>
#include <stdlib.h>
 
/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct node {
    int data;
    struct node* left;
    struct node* right;
};
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
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);
}
 
/* Given a binary tree, print its nodes according to the
  "bottom-up" postorder traversal. */
void printPostorder(struct node* node)
{
    if (node == NULL)
        return;
 
    // first recur on left subtree
    printPostorder(node->left);
 
    // then recur on right subtree
    printPostorder(node->right);
 
    // now deal with the node
    printf("%d ", node->data);
}
 
/* Driver code*/
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);
 
      // Function call
    printf("\nPostorder traversal of binary tree is \n");
    printPostorder(root);
 
    getchar();
    return 0;
}


Java




// Java program for different tree traversals
 
/* Class containing left and right child of current
   node and key value*/
class Node {
    int key;
    Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
class BinaryTree {
    // Root of Binary Tree
    Node root;
 
    BinaryTree() { root = null; }
 
    /* Given a binary tree, print its nodes according to the
      "bottom-up" postorder traversal. */
    void printPostorder(Node node)
    {
        if (node == null)
            return;
 
        // first recur on left subtree
        printPostorder(node.left);
 
        // then recur on right subtree
        printPostorder(node.right);
 
        // now deal with the node
        System.out.print(node.key + " ");
    }
 
    // Wrappers over above recursive functions
    void printPostorder() { printPostorder(root); }
 
    // Driver code
    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);
 
          // Function call
        System.out.println(
            "\nPostorder traversal of binary tree is ");
        tree.printPostorder();
    }
}


Python




# Python3 program to for tree traversals
 
# A class that represents an individual node in a
# Binary Tree
 
 
class Node:
    def __init__(self, key):
        self.left = None
        self.right = None
        self.val = key
 
# A function to do postorder tree traversal
def printPostorder(root):
 
    if root:
 
        # First recur on left child
        printPostorder(root.left)
 
        # the recur on right child
        printPostorder(root.right)
 
        # now print the data of node
        print(root.val),
 
 
# Driver code
if __name__ == "__main__":
  root = Node(1)
  root.left = Node(2)
  root.right = Node(3)
  root.left.left = Node(4)
  root.left.right = Node(5)
 
  # Function call
  print "\nPostorder traversal of binary tree is"
  printPostorder(root)


C#




// C# program for different
// tree traversals
using System;
 
/* Class containing left and
right child of current
node and key value*/
class Node {
    public int key;
    public Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
class BinaryTree {
    // Root of Binary Tree
    Node root;
 
    BinaryTree() { root = null; }
 
    /* Given a binary tree, print
       its nodes according to the
       "bottom-up" postorder traversal. */
    void printPostorder(Node node)
    {
        if (node == null)
            return;
 
        // first recur on left subtree
        printPostorder(node.left);
 
        // then recur on right subtree
        printPostorder(node.right);
 
        // now deal with the node
        Console.Write(node.key + " ");
    }
 
    // Wrappers over above recursive functions
    void printPostorder() { printPostorder(root); }
 
    // Driver Code
    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);
       
          // Function call
        Console.WriteLine("\nPostorder traversal "
                          + "of binary tree is ");
        tree.printPostorder();
    }
}
 
// This code is contributed by Arnab Kundu


Javascript




<script>
// javascript program for different tree traversals
 
/* Class containing left and right child of current
   node and key value*/
class Node {
    constructor(val) {
        this.key = val;
        this.left = null;
        this.right = null;
    }
}
 
    // Root of Binary Tree
    var root = null;
 
     
    /*
     * Given a binary tree, print its nodes according to the "bottom-up" postorder
     * traversal.
     */
    function printPostorder(node) {
        if (node == null)
            return;
 
        // first recur on left subtree
        printPostorder(node.left);
 
        // then recur on right subtree
        printPostorder(node.right);
 
        // now deal with the node
        document.write(node.key + " ");
    }
 
    /* Given a binary tree, print its nodes in inorder */
    function printInorder(node) {
        if (node == null)
            return;
 
        /* first recur on left child */
        printInorder(node.left);
 
        /* then print the data of node */
        document.write(node.key + " ");
 
        /* now recur on right child */
        printInorder(node.right);
    }
 
    /* Given a binary tree, print its nodes in preorder */
    function printPreorder(node) {
        if (node == null)
            return;
 
        /* first print data of node */
        document.write(node.key + " ");
 
        /* then recur on left subtree */
        printPreorder(node.left);
 
        /* now recur on right subtree */
        printPreorder(node.right);
        
    }
 
 
 
    // Driver method
     
     
        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);
 
        document.write("Preorder traversal of binary tree is <br/>");
        printPreorder(root);
 
        document.write("<br/>Inorder traversal of binary tree is <br/>");
        printInorder(root);
 
        document.write("<br/>Postorder traversal of binary tree is <br/>");
        printPostorder(root);
 
// This code is contributed by aashish1995
</script>


Output

Preorder traversal of binary tree is 
1 2 4 5 3 
Inorder traversal of binary tree is 
4 2 5 1 3 
Postorder traversal of binary tree is 
4 5 2 3 1 

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. 

Note: The height of the skewed tree is n (no. of elements) so the worst space complexity is O(N) and the height is (Log N) for the balanced tree so the best space
complexity is O(Log N).

Let us see different corner cases:

Complexity function T(n) — for all problems where tree traversal is involved — can be defined as: T(n) = T(k) + T(n – k – 1) + c
Where k is the number of nodes on one side of the root and n-k-1 on the other side.

Let’s do an analysis of boundary conditions:

Case 1: Skewed tree (One of the subtrees is empty and another subtree is non-empty )
k is 0 in this case. 
T(n) = T(0) + T(n-1) + c 
T(n) = 2T(0) + T(n-2) + 2c 
T(n) = 3T(0) + T(n-3) + 3c 
T(n) = 4T(0) + T(n-4) + 4c
………………………………………… 
…………………………………………. 
T(n) = (n-1)T(0) + T(1) + (n-1)c 
T(n) = nT(0) + (n)c
Value of T(0) will be some constant say d. (traversing an empty tree will take some constants time)
T(n) = n(c+d) 
T(n) = Θ(n) (Theta of n)

Case 2: Both left and right subtrees have an equal number of nodes.
T(n) = 2T(|_n/2_|) + c
This recursive function is in the standard form (T(n) = aT(n/b) + (-)(n) ) for master method. If we solve it by master method we get (-)(n)


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!