Skip to main content

Binary tree traversal in C

In an earlier post, we have seen how to add a new node to a binary tree.

Binary Tree Terminology:

  • Tree is a non-linear data structure where every node has multiple branches
  • A binary tree is a tree where each node has maximum 2 branches
  • Each node branching out is called child node 
  • The starting node of the tree is called root
  • The two children of binary tree are left child and right child.
  • The child node along with its branches and sub-branches are called sub-tree. 
  • A node has two sub-trees - left subtree and right subtree
  • A node which has no child nodes and hence no subtrees is called a leaf node
    Image from : http://msoe.us/taylor

Normally when we talk about binary tree, we refer to binary search tree which is ordered tree.
In a binary search tree, every node to the right of a given node will have larger value than the given node. And every node to the left of given node will have value smaller than given node.
That is to say the left subtree of any node has values smaller than parent and right subtree of any node has values larger than parent.

Tree traversal

In case of lists, stacks or queues, the traversal was simple. Because they are linear data structures. You start from first node, then visit the second node, then third node and so on until last node. Or you can start from last node and come backwards till first node.

But a tree is a non-linear structure. Each node has multiple branches. In case of binary tree, each node has two branches (also called children)- left and right. So which node to we visit after a given node? Its left child ? Or its right child? How do we ensure that we visit all the nodes of the tree and visit these nodes only once?

There are three ways of traversing a tree. 
  1. In order traversal - For any node, 
    1. we visit all the nodes in left subtree of a node, 
    2. visit parent,
    3. visit all nodes in right subtree
  2. Pre order traversal
    1. We visit the parent node
    2. visit all nodes of left subtree
    3. visit all nodes of right subtree.
  3. Post order traversal
    1. We visit visit all nodes of left subtree
    2. visit all nodes of right subtree
    3. and finally we visit parent node.

Let us write the inorder for the BST given above.

  1. Inorder - 1  -- 3 -- 4 -- 6 -- 7 -- 8 -- 10 -- 13 -- 14
  2. Preorder   8 -- 3 -- 1 -- 6 -- 4 -- 7 -- 10 -- 14 -- 13
  3. Postorder  1 -- 4 -- 7 -- 6 -- 3 -- 13 -- 14 -- 10 -- 8

In inorder, we start with root 8. But before we visit 8, we should visit all nodes of left subtree. So we branch to 3. Before we visit 3, we go to its left child 1. Since 1 has no left child, we print 1. Now we have visited complete left branch of 3. So we print 3 and then we go to 6. Before we print 6, we should go to its left child 4. As 4 is leaf node (node with no children), it is printed. After visiting 4, which is left branch of 6, we can print parent viz 6. Next we go to right branch of 6.

 So you can write down like this.

But a recursive function for these in very simple. Or really very small.



void in_order(NODEPTR nd)
{
   if(nd!=NULL)
    {
        in_order(nd->left);/*visit left subtree*/
        printf("%d---",nd->val);
        in_order(nd->right);/*visit right subtree*/
     }
}

When you run the program, you will notice that, inorder traversal will print the nodes in ascending order of values.

Similarly you can write preorder and postorder traversal functions too.


void pre_order(NODEPTR nd)
{
   if(nd!=NULL)
    {
        printf("%d---",nd->val);   
        pre_order(nd->left);
        pre_order(nd->right);
            
     }
}

void post_order(NODEPTR nd)
{
   if(nd!=NULL)
    {    
 post_order(nd->left);
        post_order(nd->right);
        printf("%d---",nd->val);
     }
}

Here is the complete program.

#include<stdio.h>
#include<stdlib.h>
struct node
{
   int val;
   struct node *left;
   struct node *right;
};
typedef struct node *NODEPTR;

NODEPTR create_node(int num)
{
     NODEPTR temp = (NODEPTR)malloc(sizeof(struct node));
     temp->val = num;
     temp->left = NULL;
     temp->right = NULL;
     return temp;
}

NODEPTR insert_node(NODEPTR nd,NODEPTR newnode)
{
    if(nd==NULL)
       return newnode;/* newnode becomes root of tree*/
    if(newnode->val > nd->val)
        nd->right = insert_node(nd->right,newnode);
    else if(newnode->val <  nd->val)
        nd->left = insert_node(nd->left,newnode); 
    return nd;   
}

void in_order(NODEPTR nd)
{
   if(nd!=NULL)
    {
        in_order(nd->left);
        printf("%d---",nd->val);
        in_order(nd->right);
     }
}

void pre_order(NODEPTR nd)
{
   if(nd!=NULL)
    {
        printf("%d---",nd->val);   
        pre_order(nd->left);
        pre_order(nd->right);
            
     }
}

void post_order(NODEPTR nd)
{
   if(nd!=NULL)
    {    
 post_order(nd->left);
        post_order(nd->right);
        printf("%d---",nd->val);
     }
}

int main()
{
       NODEPTR root=NULL; 
       int n;
       do
       {
           NODEPTR newnode;
           printf("Enter value of node(-1 to exit):");
           scanf("%d",&n);
           if(n!=-1)
            {  
               newnode = create_node(n);
               root = insert_node(root,newnode);
             }
       } while (n!=-1);
       printf("Preorder traversal\n");
       pre_order(root);
       printf("\nInorder traversal\n");
       in_order(root);
       printf("\nPostorder traversal\n");
       post_order(root);
       return 0;
}

Comments

Popular posts from this blog

Introduction to AVL tree

AVL tree is a balanced binary search tree where the difference between heights of two sub trees is maximum 1. Why balanced tree A binary tree is good data structure because search operation here is of the order of O(logn). But this is true if the tree is balanced - which means the left and right subtrees are almost equal in height. If not balanced, search operation will take longer.  In worst case, if the tree has only one branch, then search is of the order O(n). Look at this example.  Here all nodes have only right children.  To search a value in this tree, we need upto 7 iterations, which is O(n). So this tree is very very inefficient. One way of making the tree efficient is to, balance the tree and make sure that height of two branches of each node are almost equal. Height of a node Height of a node is the distance between the node and its extreme child.  In the above a diagram, height of 37 is 3 and height of left child of 37 is 0 and right child of 37 is 2. Bal...

Balanced brackets

Have you observed something? When ever you are writing code using any IDE, if you write mismatched brackets, immediately an error is shown by IDE. So how does IDE  know if an expression is having balanced brackets? For that, the expression must have equal number of opening and closing brackets of matching types and also they must be in correct order. Let us look at some examples (a+b)*c+d*{e+(f*g)}   - balanced (p+q*[r+u )] - unbalanced (p+q+r+s) ) - unbalanced (m+n*[p+q]+{g+h}) - balanced So we do we write a program to check if an expression is having balanced brackets? We do need to make use of stack to store the brackets. The algorithm is as follows Scan a character - ch from the expression If the character is opening bracket, push it to stack If the character is closing bracket pop a character from stack If popped opening bracket and ch are not of same type ( ( and ) or [ and ] ) stop the function and return false Repeat steps 2 and 3 till all characters are scanned. On...

Program to delete a node from linked list

How do you remove a node from a linked list? If you have to delete a node, first you need to search the node. Then you should find its previous node. Then you should link the previous node to the next node. If node containing 8 has to be deleted, then n1 should be pointing to n2. Looks quite simple. Isn't it? But there are at least two special cases you have to consider. Of course, when the node is not found. If the node is first node of the list viz head. If the node to be deleted is head node, then if you delete, the list would be lost. You should avoid that and make the second node as the head node. So it becomes mandatory that you return the changed head node from the function.   Now let us have a look at the code. #include<stdio.h> #include<stdlib.h> struct node { int data; struct node * next; }; typedef struct node * NODEPTR; NODEPTR create_node ( int value) { NODEPTR temp = (NODEPTR) malloc( size...