Inorder Sorting of a Binary Tree to an Array

Question

I am trying to traverse a Binary Tree in descending inorder style and copy the elements to an array. I believe I am 80% there, I just can't seem to figure out what's going on with my index. Any help or pointers in the right direction are appreciated.

public static void inorder(int[] a) {
    int i = 0;

    // call recursion
    inorder(root, a, i);


    System.out.println("Array: ");
    for (int j = 0; j < a.length; j++) {
        System.out.println(a[j]);
    }
}

private static void inorder(Node temp, int[] a, int i) {
    // base case
    if (temp == null) return;

    // go to the right of tree
    inorder(temp.right, a, i);


    // copy node to array
    a[i] = temp.number;
    System.out.println(temp.number);
    System.out.println(i);
    i++;

    // go to the left of tree
    inorder(temp.left, a, i);
}

Normally yes, but that would pull the nodes in ascending order. I'm going for descending. — kcl1924
– kcl1924, Commented Dec 11, 2014 at 9:32
I believe that the increments done to i in recursive calls do not modify it's value at the current level. Try using Integer i instead of int i in your method declaration. That might help. — Dušan Rychnovský
– Dušan Rychnovský, Commented Dec 11, 2014 at 9:42
@DušanRychnovský I get the same result. Index "i" seems to stay in the lower region of indices. If I insert 10 values into the tree and try to sort them, "i" never passes 3. — kcl1924
– kcl1924, Commented Dec 11, 2014 at 9:51

softwarenewbie7331 · Accepted Answer · 2014-12-11 09:50:00Z

1

try updating i and returning its value

private static int inorder(Node temp, int[] a, int i) {
// base case
if (temp == null) return i;

// go to the right of tree
i = inorder(temp.right, a, i);


// copy node to array
a[i] = temp.number;
System.out.println(temp.number);
System.out.println(i);
i++;

// go to the left of tree
i = inorder(temp.left, a, i);
return i;
}

answered Dec 11, 2014 at 9:50

softwarenewbie7331

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