Convert Sorted Array to Binary Search Tree [duplicate]

Question

I am working on "Convert Sorted Array to Binary Search Tree With Minimal Height", which asked:

Given a sorted (increasing order) array, Convert it to create a binary tree with minimal height.

I am not able to find why my recursive does not stop as I expected. It should stop when 7 passed, and will not print out 7 again. I also found a similar answer, it looks like used same strategy as mine, but it works fine. (I don't think my question is duplicate as those questions listed above, but I still want to thank you to link them for me. They gave me more idea to solve my problem.)

My code is below:

public TreeNode sortedArrayToBST(int[] A) {  
    int len = A.length;
    if(len <= 0){
        return null;
    }

    TreeNode root = new TreeNode(A[(len - 1) / 2]);
    if(len == 1){
        return root;
    }
    else{
        helper(root, A, 0, len - 1);
    }
    return root;
}

public void helper(TreeNode root, int[] A, int leftPoint, int rightPoint){
    if((rightPoint - leftPoint) <= 0){
        return;
    }

    int mid = (rightPoint - leftPoint) / 2 + leftPoint;
    int leftChild = (mid - 1 - leftPoint) / 2 + leftPoint;
    int rightChild = (rightPoint - (mid + 1)) / 2 + mid + 1;

    TreeNode left = new TreeNode(A[leftChild]);
    root.left = left;

    TreeNode right = new TreeNode(A[rightChild]);
    root.right = right;

    helper(root.left, A, leftPoint, mid - 1);
    helper(root.right, A, mid + 1, rightPoint);
    return;
}

When I run it, I got this.

My input
[1,2,3,4,5,6,7,8]

My output
{4,2,6,1,3,5,7,#,#,#,#,#,#,7,8}

Expected
{4,2,6,1,3,5,7,#,#,#,#,#,#,#,8}

Why does it have duplicate 7 at right side? As 7 has been used, it should be kicked out.

And I found my idea is similar with the following answer:

public TreeNode sortedArrayToBST(int[] A) {  
    // write your code here
    int len = A.length;
    if(len <= 0){
        return null;
    }
    TreeNode root = helper1(A, 0, len - 1);
    return root;
}

public TreeNode helper1(int[] A, int low, int high){
    if(low > high){
        return null;
    }
    int mid = (high + low) / 2;
    TreeNode root = new TreeNode(A[mid]);
    root.left = helper1(A, low, mid - 1);
    root.right = helper1(A, mid + 1, high);
    return root;
}

Answer 1

Let's have the following array:

[1,2,3,4,5,6,7]

The expected BST is:

       4
    2     6
  1   3  5  7

To achieve that, we can go in the following way:

for (int i = 0; i < logn; i++) {
    //insert ith level
}

To make things easier, lets find min n, so n > array.length and n = 2^k.

On ith level, starting from i = 0 we've got:

n/2^(i+1), 3*n/2^(i+1), 5*n/2^(i+1)...

The numbers above, are all indexes in the array.

public TreeNode sortedArrayToBST(int[] A) {  
    int len = A.length;
    if(len <= 0){
        return null;
    }

    int n = 1;
    int i = 0;
    while (n < len) {
        n *= 2;
        i++;
    }

    TreeNode root = new TreeNode(A[n/2]);

    for (int j = 1; j < i; j++) {
        insert(root, j, n, A);
    }
}

private void insert(TreeNode root, int j, int n, int[] A) {

    int helper = n/Math.pow(2, j+1);
    for (int i = 1; i <= Math.pow(2, j); i ++) {
        root.add(A[i*helper]);
    }
}