- Write a C program to print level of a node in binary tree.
- Recursive function in C to find level of a given node in binary tree.

Given a binary tree and a node N, we have to **print the level of node N in binary tree**. To **print the level of a node** N, first of all we have to search it in binary tree. Here, we can use In Order, pre Order or post Order traversal to search node N in binary tree. We will also keep track of level of current node while traversing.

**Algorithm to find level of a given node in binary tree**

- Traverse given binary tree using pre order traversal by keeping track of levels of nodes.
- Let node be the pointer to any node at level L.
- If node is equal to NULL, return.
- If node is equal to N, then print the level of node(L) on screen else continue traversal of sub trees at level L+1.

**Time Complexity**: O(n), we are traversing binary tree only once.

## C program to print level of a given node of binary tree

#include <stdio.h> struct node { int data; struct node *left; struct node *right; }; struct node* getNewNode(int data) { /* dynamically allocate memory for a new node */ struct node* newNode = (struct node*)malloc(sizeof(struct node)); /* populate data in new Node */ newNode->data = data; newNode->left = NULL; newNode->right = NULL; return newNode; } /* This function returns below tree 1 / \ 2 3 / \ / \ 4 5 6 7 / \ \ 8 9 10 */ struct node* generateBTree(){ // Root Node struct node* root = getNewNode(1); root->left = getNewNode(2); root->right = getNewNode(3); root->left->left = getNewNode(4); root->left->right = getNewNode(5); root->right->left = getNewNode(6); root->right->right = getNewNode(7); root->left->left->left = getNewNode(8); root->left->left->right = getNewNode(9); root->right->right->right = getNewNode(10); return root; } /* Prints level of all nodes. It does pre Order traversal and keeps track of the current level and prints it. */ void printLevelOfNode(struct node* root, int currentLevel, int num) { if(root == NULL) { return; } if(root->data == num) { printf("Level of %d is %d \n", num, currentLevel); } printLevelOfNode(root->left, currentLevel+1, num); printLevelOfNode(root->right, currentLevel+1, num); } int main() { struct node *root = generateBTree(); /*Printing level of nodes */ printLevelOfNode(root, 0, 1); printLevelOfNode(root, 0, 5); printLevelOfNode(root, 0, 7); printLevelOfNode(root, 0, 9); getchar(); return 0; }Output

Level of 1 is 0 Level of 5 is 2 Level of 7 is 2 Level of 9 is 3