IBM Interview Question for Software Engineer / Developers


Country: United States
Interview Type: Phone Interview




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1
of 1 vote

from binaryTree import BinaryTree

def arrayBST(l,start,end):
	mid=(start+end)/2
	if start<=end:
		tree=BinaryTree(l[mid])
		tree.left=arrayBST(l,start,mid-1)
		tree.right=arrayBST(l,mid+1,end)
		return tree

def inorder(tree):
	if tree!=None:
		inorder(tree.left)
		print tree.key
		inorder(tree.right)

def preorder(tree):
	if tree!=None:
		print tree.key
		preorder(tree.left)
		preorder(tree.right)

t=arrayBST([1,2,3,4],0,3)

inorder(t)
print 
preorder(t)

- Yash Katariya January 18, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

Suppose we need to build a balanced binary search tree T for the elements in sorted array A, from index "lo" to "hi".

build(A, lo, hi):
The root of T must be the middle element: T.root = A[mid], where mid = (lo+hi)/2

The left subtree of T must be built from A[lo, mid), recursively:
T.left = build(A, lo, mid-1)

The right subtree of T must be built from A[mid+1, hi), recursively:
T.right = build(A, mid+1, hi)

Remember the basecase is when hi-lo <=1.

Complexity:
Since each element of the array A is accessed once, the complexity is O(N), N = length of array A.

- ninhnnsoc January 18, 2016 | Flag Reply
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0
of 0 vote

TreeNode* BuildMinimalTreeFromArr(int* arr , int left, int right){
if(left > right){
return NULL;
}
else{
int mid = (left+right)/2;
TreeNode* root = new TreeNode;
root->data = arr[mid];
root->left = BuildMinimalTreeFromArr(arr,left,mid-1);
root->right = BuildMinimalTreeFromArr(arr,mid+1,right);
return root;
}

}

- zen January 27, 2016 | Flag Reply
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0
of 0 vote

Building balanced BST from sorted array (Swift):

class Node {
    var data: Int
    var left: Node?
    var right: Node?
    
    init(data: Int, left: Node?, right: Node?) {
        self.data = data
        self.left = left
        self.right = right
    }
}

var arr = [1,2,3,4,5,6,7,8,9]

func buildTree(arr: [Int]) -> Node? {
    if arr.count == 0 {
        return nil
    }
    
    var mid = arr.count / 2
    
    var left : Node? = nil
    if mid > 0 {
        left = buildTree(Array(arr[0..<mid]))
    }
    var right : Node? = nil
    if mid < arr.count - 1 {
        right = buildTree(Array(arr[mid+1..<arr.count]))
    }
    let node = Node(data: arr[mid], left: left, right: right)
    return node
}

var root = buildTree(arr)

- aquio July 19, 2016 | Flag Reply
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0
of 0 vote

The iterative solution -

class TreeNode {
	int val;
	TreeNode left;
	TreeNode right;
	
	public TreeNode(int val){
		this.val = val;
	}
}

public static TreeNode buildTreeIterative(int[] nums) {
		Stack<int[]> pos = new Stack<>();
		Stack<TreeNode> nodes = new Stack<>();
		
		TreeNode root = new TreeNode(-1);
		
		pos.push(new int[] {0, nums.length - 1});
		nodes.push(root);
		
		while(!nodes.isEmpty()) {
			// Get the current node
			TreeNode node = nodes.pop();
			
			// Pop an element from the stack
			int[] range = pos.pop();
			int start = range[0];
			int end = range[1];
			int mid = range[0] + (range[1] - range[0])/2;
			node.val = nums[mid];
			
			if(start < mid) {
				// There exists a left child for the node
				node.left = new TreeNode(-1);
				pos.push(new int[] {start, mid - 1});
				nodes.push(node.left);
			}
			
			if(end > mid) {
				// There exists a right child for the node
				node.right = new TreeNode(-1);
				pos.push(new int[] {mid + 1, end});
				nodes.push(node.right);
			}
		}
		
		return root;
	}

- anshul.chandra October 07, 2017 | Flag Reply


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