How to keep track of layers when traversing a binary tree? - c++

Question

If I need to print out each elements of a binary tree constructed with the struct below. How could I keep track of which layer of elements I am printing?
struct for a binary tree node
For example:
any binary tree
Expected output:
layer 0: 12
layer -1: 28 19
layer -2: 94 32
layer -3: 65 18 72

Answer 1

Solution using queue based on GeeksForGeeks
#include <iostream>
#include <queue>
using namespace std;
// A Binary Tree Node
struct node
{
struct node *left;
int data;
struct node *right;
};
// Iterative method to do level order traversal
// line by line
void printLevelOrder(node *root)
{
// Base Case
if (root == NULL)
return;
// Create an empty queue for level order tarversal
queue<node *> q;
// Enqueue Root and initialize height
q.push(root);
int i = 0;
while (q.empty() == false)
{
cout << "layer " << i << ": ";
// nodeCount (queue size) indicates number
// of nodes at current lelvel.
int nodeCount = q.size();
// Dequeue all nodes of current level and
// Enqueue all nodes of next level
while (nodeCount > 0)
{
node *node = q.front();
cout << node->data << " ";
q.pop();
if (node->left != NULL)
q.push(node->left);
if (node->right != NULL)
q.push(node->right);
nodeCount--;
}
cout << endl;
--i;
}
}
// Utility function to create a new tree node
node *newNode(int data)
{
node *temp = new node;
temp->data = data;
temp->left = NULL;
temp->right = NULL;
return temp;
}
// Driver program to test above functions
int main()
{
// Create binary tree
node *root = newNode(12);
root->left = newNode(28);
root->right = newNode(19);
root->left->left = newNode(94);
root->left->left->left = newNode(65);
root->left->left->right = newNode(18);
root->right->left = newNode(32);
root->right->left->right = newNode(72);
printLevelOrder(root);
return 0;
}
Solution using recursive function and helper function based on CrazyForCode:
#include <iostream>
using namespace std;
struct node
{
int data;
struct node *left;
struct node *right;
};
void printLevel(node *, int);
int height(struct node *node);
/* Function to print level order traversal a tree*/
void printLevelOrder(struct node *root)
{
int h = height(root);
int i;
for (i = 1; i <= h; i++){
printf("layer %d: ",i*-1+1);
printLevel(root, i);
cout << endl;
}
}
/* Print nodes at a given level */
void printLevel(struct node *root, int level)
{
if (root == NULL)
return;
if (level == 1)
{
printf("%d ", root->data);
}
else if (level > 1)
{
printLevel(root->left, level - 1);
printLevel(root->right, level - 1);
}
}
/* Compute the "height" of a tree */
int height(struct node *node)
{
if (node == NULL)
return 0;
else
{
int lheight = height(node->left);
int rheight = height(node->right);
if (lheight > rheight)
return (lheight + 1);
else
return (rheight + 1);
}
}
node *newNode(int data)
{
node *temp = new node;
temp->data = data;
temp->left = NULL;
temp->right = NULL;
return temp;
}
int main()
{
// Create binary tree
node *root = newNode(12);
root->left = newNode(28);
root->right = newNode(19);
root->left->left = newNode(94);
root->left->left->left = newNode(65);
root->left->left->right = newNode(18);
root->right->left = newNode(32);
root->right->left->right = newNode(72);
printLevelOrder(root);
return 0;
}