Given a Binary tree, Traverse it using DFS using recursion.
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. Generally there are 2 widely used ways for traversing trees:
- DFS or Depth First Search
- BFS or Breadth First Search
In this article, traversal using DFS has been discussed. Please see this post for Breadth First Traversal.
Depth-first search
DFS (Depth-first search) is technique used for traversing tree or graph. Here backtracking is used for traversal. In this traversal first the deepest node is visited and then backtracks to it’s parent node if no sibling of that node exist.
DFS Traversal of a Graph vs Tree
In graph, there might be cycles and dis-connectivity. Unlike graph, tree does not contain cycle and always connected. So DFS of a tree is relatively easier. We can simply begin from a node, then traverse its adjacent (or children) without caring about cycles. And if we begin from a single node (root), and traverse this way, it is guaranteed that we traverse the whole tree as there is no dis-connectivity,
Examples:
Tree:Therefore, the Depth First Traversals of this Tree will be: (a) Inorder (Left, Root, Right) : 4 2 5 1 3 (b) Preorder (Root, Left, Right) : 1 2 4 5 3 (c) Postorder (Left, Right, Root) : 4 5 2 3 1Example Tree
Below are the Tree traversals through DFS using recursion:
-
Inorder Traversal (Practice):
Example: Inorder traversal for the above-given figure is 4 2 5 1 3.
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)
Implementation:
C++
// C program for different tree traversals
#include <iostream>
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;
Node(
int
data)
{
this
->data = data;
left = right = NULL;
}
};
/* 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 program to test above functions*/
int
main()
{
struct
Node* 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);
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 program to test above functions*/
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
// 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 method
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
# Python 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
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#
// 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
;
}
}
public
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
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);
Console.WriteLine(
"\nInorder traversal of binary tree is "
);
tree.printInorder();
}
}
// This code is contributed by PrinciRaj1992
Output:Inorder traversal of binary tree is 4 2 5 1 3
Uses of Inorder:
In 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 s reversed can be used. -
Preorder Traversal (Practice):
Example: Preorder traversal for the above given figure is 1 2 4 5 3.
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)
Implementation:
C++
// C program for different tree traversals
#include <iostream>
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;
Node(
int
data)
{
this
->data = data;
left = right = NULL;
}
};
/* 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 sutree */
printPreorder(node->left);
/* now recur on right subtree */
printPreorder(node->right);
}
/* Driver program to test above functions*/
int
main()
{
struct
Node* 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);
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 sutree */
printPreorder(node->left);
/* now recur on right subtree */
printPreorder(node->right);
}
/* Driver program to test above functions*/
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
// 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 sutree */
printPreorder(node.left);
/* now recur on right subtree */
printPreorder(node.right);
}
// Wrappers over above recursive functions
void
printPreorder() { printPreorder(root); }
// Driver method
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
# Python 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
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#
// C# program for different tree traversals
using
System;
/* Class containing left and right child of current
node and key value*/
public
class
Node
{
public
int
key;
public
Node left, right;
public
Node(
int
item)
{
key = item;
left = right =
null
;
}
}
public
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 sutree */
printPreorder(node.left);
/* now recur on right subtree */
printPreorder(node.right);
}
// Wrappers over above recursive functions
void
printPreorder() { printPreorder(root); }
// Driver method
public
static
void
Main()
{
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();
}
}
/* This code contributed by PrinciRaj1992 */
Output:Preorder traversal of binary tree is 1 2 4 5 3
Uses of Preorder:
Preorder traversal is used to create a copy of the tree. Preorder traversal is also used to get prefix expression on of an expression tree. Please see http://en.wikipedia.org/wiki/Polish_notation to know why prefix expressions are useful. -
Postorder Traversal (Practice):
Example: Postorder traversal for the above given Tree is 4 5 2 3 1.
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.
Implementation:
C++
// C program for different tree traversals
#include <iostream>
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;
Node(
int
data)
{
this
->data = data;
left = right = NULL;
}
};
/* 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 program to test above functions*/
int
main()
{
struct
Node* 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);
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 program to test above functions*/
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
// 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 method
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
# Python 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
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#
// C# program for different tree traversals
using
System;
/* Class containing left and right child of current
node and key value*/
public
class
Node
{
public
int
key;
public
Node left, right;
public
Node(
int
item)
{
key = item;
left = right =
null
;
}
}
public
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
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);
Console.WriteLine(
"\nPostorder traversal of binary tree is "
);
tree.printPostorder();
}
}
// This code contributed by Rajput-Ji
Output:Postorder traversal of binary tree is 4 5 2 3 1
Uses of Postorder:
Postorder traversal is used to delete the tree. Please see the question for deletion of tree for details. Postorder traversal is also useful to get the postfix expression of an expression tree. Please see http://en.wikipedia.org/wiki/Reverse_Polish_notation to for the usage of postfix expression.
Implementing all traversals using DFS
C++
// C program for different tree traversals #include <iostream> 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; Node( int data) { this ->data = data; left = right = NULL; } }; /* 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 << " " ; } /* 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); } /* 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 sutree */ printPreorder(node->left); /* now recur on right subtree */ printPreorder(node->right); } /* Driver program to test above functions*/ int main() { struct Node *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); cout << "\nPreorder traversal of binary tree is \n" ; printPreorder(root); cout << "\nInorder traversal of binary tree is \n" ; printInorder(root); 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); } /* 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); } /* 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 sutree */ printPreorder(node->left); /* now recur on right subtree */ printPreorder(node->right); } /* Driver program to test above functions*/ 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); printf ( "\nInorder traversal of binary tree is \n" ); printInorder(root); 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 + " " ); } /* 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); } /* 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 sutree */ printPreorder(node.left); /* now recur on right subtree */ printPreorder(node.right); } // Wrappers over above recursive functions void printPostorder() { printPostorder(root); } void printInorder() { printInorder(root); } void printPreorder() { printPreorder(root); } // Driver method 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(); System.out.println( "\nInorder traversal of binary tree is " ); tree.printInorder(); System.out.println( "\nPostorder traversal of binary tree is " ); tree.printPostorder(); } } |
Python
# Python 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) # 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), # 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 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) print "\nInorder traversal of binary tree is" printInorder(root) print "\nPostorder traversal of binary tree is" printPostorder(root) |
C#
// C# program for different Console.Writetree traversals using System; /* Class containing left and right child of current node and key value*/ public class Node { public int key; public Node left, right; public Node( int item) { key = item; left = right = null ; } } public 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 + " " ); } /* 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); } /* 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 sutree */ printPreorder(node.left); /* now recur on right subtree */ printPreorder(node.right); } // Wrappers over above recursive functions void printPostorder() { printPostorder(root); } void printInorder() { printInorder(root); } 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); Console.WriteLine( "Preorder traversal of binary tree is " ); tree.printPreorder(); Console.WriteLine( "\nInorder traversal of binary tree is " ); tree.printInorder(); Console.WriteLine( "\nPostorder traversal of binary tree is " ); tree.printPostorder(); } } // This code has been contributed by 29AjayKumar |
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 size of stack for function calls then O(1) otherwise O(n).
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