Question

This is a question I thought it would be easy but I found I’m wrong in the last. I can finish the program without recursive but I want to ask whether this problem can be finished in recursive version or not?

Answer 1

A recursive binary search tree traversal is basically (pseudo-code in case this is coursework):

def traverse (node):
    if (node == NULL):
        return
    traverse (node.left)
    doSomethingWith (node.payload)
    traverse (node.right)
:
traverse (root)

That’s all there is to it really, just replace doSomethingWith() with whatever you want to do (such as print).

That will traverse in left to right order so, if your BST is ordered in such a way that left means lower, simply swap over the two traverse calls.

By way of example, consider the following tree:

       20
      /  \
    10    25
   /     /  \
  5    24    27
 /          /
2         28

as embodied in this example C program:

#include <stdio.h>

typedef struct s {
    int payload;
    int left;
    int right;
} tNode;

tNode node[] = {  // Trust me, this is the tree from above :-)
    {20,  1,  4}, {10,  2, -1}, { 5,  3, -1}, { 2, -1, -1},
    {25,  5,  6}, {24, -1, -1}, {27, -1,  7}, {28, -1, -1}};

static void traverse (int idx) {
    if (idx == -1) return;
    traverse (node[idx].right);
    printf ("%d ", node[idx].payload);
    traverse (node[idx].left);
}

int main (void) {
    traverse (0);
    putchar ('\n');
    return 0;
}

Running that program gives you the following output:

28 27 25 24 20 10 5 2