Print level and nodes from a binary search tree

Question

I need to do an inorder traversal of a binary search tree, and I need to print all the nodes and the level they're at, but I can't think of a way to do this.

For example:

If I have this bst , the output would be:

4 #1
5 #0
9 #1
7 #2
10 #2
18 #3

This is so far what I got:

This is the struct I'm using:

struct tree {
    int number;
    tree *izq;
    tree *der;
};

typedef struct tree *bst;

And this is the function I'm trying to implement:

void printTree(bst node) {
    if (node==NULL) {
        return;
    }
    else {
        if (node->left != NULL) {
            printTree(node->left);
        }
        printf("%d", node->number);
        if (node->right !=NULL) {
            printTree(node->right);
        }
    }
}

Does anyone have any ideas? Thank you :)

Answer 1

I implemented this with Java, but I think you can easily convert it to C:

private static void printWithLevels(TreeNode node) {
    printWithLevels(node, 0);
}

private static void printWithLevels(TreeNode node, int level) {
    if (node == null) return;

    System.out.println(node.value + "(" + level + ")");

    printWithLevels(node.left, level + 1);
    printWithLevels(node.right, level + 1);
}

For my solution to be complete, this is my simple/quick implementation of TreeNode:

private static class TreeNode {
    int value;
    TreeNode left;
    TreeNode right;

    TreeNode(int value, TreeNode left, TreeNode right) {
        this.value = value;
        this.left = left;
        this.right = right;
    }
}

Answer 2

warning

struct tree {
    int number;
    tree *izq;
    tree *der;
};

must be

struct tree {
    int number;
    struct tree *izq;
    struct tree *der;
};

---

because the case where node is NULL is checked at the beginning you can simplify :

void printTree(bst node) {
    if (node != NULL) {
      printTree(node->izq);
      printf("%d", node->number);
      printTree(node->der);
    }
}

adding the level:

void printTree(bst node, int lvl) {
    if (node != NULL) {
      printTree(node->izq, lvl + 1);
      printf("%d #%d\n", node->number, lvl);
      printTree(node->der, lvl + 1);
    }
}

and you call at the root level with level 0

---

Making a full program :

#include <stdio.h>
#include <stdlib.h>

struct tree {
    int number;
    struct tree *izq;
    struct tree *der;
};

typedef struct tree *bst;

void printTree(bst node, int lvl) {
    if (node != NULL) {
      printTree(node->izq, lvl + 1);
      printf("%d #%d\n", node->number, lvl);
      printTree(node->der, lvl + 1);
    }
}

struct tree * mk(int v, struct tree * l, struct tree * r)
{
  struct tree * t = malloc(sizeof(struct tree));

  t->number = v;
  t->izq = l;
  t->der = r;

  return t;
}

int main()
{
  struct tree * r = mk(5, mk(4, NULL, NULL), mk(9, mk(7, NULL, NULL), mk(10, NULL, mk(18, NULL, NULL))));

  printTree(r, 0);
}

Compilation an execution :

pi@raspberrypi:/tmp $ gcc -pedantic -Wall -Wextra c.c
pi@raspberrypi:/tmp $ ./a.out
4 #1
5 #0
7 #2
9 #1
10 #2
18 #3