draw an expression tree for the inorder traversal
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)
- Traverse the left subtree, i.e., call Inorder(left->subtree)
- Visit the root.
- 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++
#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 << "\nInorder traversal of binary tree is \n" ;
printInorder(root);
return 0;
}
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 ( "\nInorder traversal of binary tree is \n" );
printInorder(root);
getchar ();
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);
}
void printInorder() { printInorder(root); }
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(
"\nInorder traversal of binary tree is " );
tree.printInorder();
}
}
Python
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),
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 "\nInorder 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( "\nInorder traversal "
+ "of binary tree is " );
tree.printInorder();
}
}
Javascript
<script>
class Node {
constructor(val) {
this .key = val;
this .left = null ;
this .right = null ;
}
}
var root = null ;
function printPostorder(node) {
if (node == null )
return ;
printPostorder(node.left);
printPostorder(node.right);
document.write(node.key + " " );
}
function printInorder(node) {
if (node == null )
return ;
printInorder(node.left);
document.write(node.key + " " );
printInorder(node.right);
}
function printPreorder(node) {
if (node == null )
return ;
document.write(node.key + " " );
printPreorder(node.left);
printPreorder(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);
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);
</script>
Preorder Traversal ( Practice ):
Algorithm Preorder(tree)
- Visit the root.
- Traverse the left subtree, i.e., call Preorder(left->subtree)
- 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++
#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 printPreorder( struct Node* node)
{
if (node == NULL)
return ;
cout << 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);
cout << "\nPreorder traversal of binary tree is \n" ;
printPreorder(root);
return 0;
}
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 ( "\nPreorder traversal of binary tree is \n" );
printPreorder(root);
getchar ();
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 printPreorder(Node node)
{
if (node == null )
return ;
System.out.print(node.key + " " );
printPreorder(node.left);
printPreorder(node.right);
}
void printPreorder() { printPreorder(root); }
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(
"Preorder traversal of binary tree is " );
tree.printPreorder();
}
}
Python
class Node:
def __init__( self , key):
self .left = None
self .right = None
self .val = key
def printPreorder(root):
if root:
print (root.val),
printPreorder(root.left)
printPreorder(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 "Preorder traversal of binary tree is"
printPreorder(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 printPreorder(Node node)
{
if (node == null )
return ;
Console.Write(node.key + " " );
printPreorder(node.left);
printPreorder(node.right);
}
void printPreorder() { printPreorder(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( "Preorder traversal "
+ "of binary tree is " );
tree.printPreorder();
}
}
Javascript
<script>
class Node {
constructor(val) {
this .key = val;
this .left = null ;
this .right = null ;
}
}
var root = null ;
function printPostorder(node) {
if (node == null )
return ;
printPostorder(node.left);
printPostorder(node.right);
document.write(node.key + " " );
}
function printInorder(node) {
if (node == null )
return ;
printInorder(node.left);
document.write(node.key + " " );
printInorder(node.right);
}
function printPreorder(node) {
if (node == null )
return ;
document.write(node.key + " " );
printPreorder(node.left);
printPreorder(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);
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);
</script>
Postorder Traversal ( Practice ):
Algorithm Postorder(tree)
- Traverse the left subtree, i.e., call Postorder(left->subtree)
- Traverse the right subtree, i.e., call Postorder(right->subtree)
- 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++
#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 printPostorder( struct Node* node)
{
if (node == NULL)
return ;
printPostorder(node->left);
printPostorder(node->right);
cout << node->data << " " ;
}
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 << "\nPostorder traversal of binary tree is \n" ;
printPostorder(root);
return 0;
}
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 printPostorder( struct node* node)
{
if (node == NULL)
return ;
printPostorder(node->left);
printPostorder(node->right);
printf ( "%d " , node->data);
}
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 ( "\nPostorder traversal of binary tree is \n" );
printPostorder(root);
getchar ();
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 printPostorder(Node node)
{
if (node == null )
return ;
printPostorder(node.left);
printPostorder(node.right);
System.out.print(node.key + " " );
}
void printPostorder() { printPostorder(root); }
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(
"\nPostorder traversal of binary tree is " );
tree.printPostorder();
}
}
Python
class Node:
def __init__( self , key):
self .left = None
self .right = None
self .val = key
def printPostorder(root):
if root:
printPostorder(root.left)
printPostorder(root.right)
print (root.val),
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 "\nPostorder traversal of binary tree is"
printPostorder(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 printPostorder(Node node)
{
if (node == null )
return ;
printPostorder(node.left);
printPostorder(node.right);
Console.Write(node.key + " " );
}
void printPostorder() { printPostorder(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( "\nPostorder traversal "
+ "of binary tree is " );
tree.printPostorder();
}
}
Javascript
<script>
class Node {
constructor(val) {
this .key = val;
this .left = null ;
this .right = null ;
}
}
var root = null ;
function printPostorder(node) {
if (node == null )
return ;
printPostorder(node.left);
printPostorder(node.right);
document.write(node.key + " " );
}
function printInorder(node) {
if (node == null )
return ;
printInorder(node.left);
document.write(node.key + " " );
printInorder(node.right);
}
function printPreorder(node) {
if (node == null )
return ;
document.write(node.key + " " );
printPreorder(node.left);
printPreorder(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);
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);
</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)
Source: https://www.geeksforgeeks.org/tree-traversals-inorder-preorder-and-postorder/
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