Interview Question

Data Scientist Intern Interview

-Mountain View, CA

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Find the second largest element in a Binary Search Tree

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Interview Answers

16 Answers

14

The above answer is also wrong; Node findSceondLargest(Node root) { // If tree is null or is single node only, return null (no second largest) if (root==null || (root.left==null && root.right==null)) return null; Node parent = null, child = root; // find the right most child while (child.right!=null) { parent = child; child = child.right; } // if the right most child has no left child, then it's parent is second largest if (child.left==null) return parent; // otherwise, return left child's rightmost child as second largest child = child.left; while (child.right!=null) child = child.right; return child; }

mindpower on

25

find the right most element. If this is a right node with no children, return its parent. if this is not, return the largest element of its left child.

Anonymous on

5

One addition is the situation where the tree has no right branch (root is largest). In this special case, it does not have a parent. So it's better to keep track of parent and current pointers, if different, the original method by the candidate works well, if the same (which means the root situation), find the largest of its left branch.

Anonymous on

2

int getmax(node *root) { if(root->right == NULL) { return root->d; } return getmax(root->right); } int secondmax(node *root) { if(root == NULL) { return -1; } if(root->right == NULL && root->left != NULL) { return getmax(root->left); } if(root->right != NULL) { if(root->right->right == NULL && root->right->left == NULL) { return root->d; } } return secondmax(root->right); }

Anonymous on

2

Above answer is wrong. it has to be something like this. public static int findSecondLargest(Node node) { Node secondLargest = null; Node parent = null; Node child = node; if (node!=null && (node.hasLeftChild()||node.hasRightChild())) { if (node.hasRightChild()) { while (child.hasRightChild()) { parent = child; child = child.rightChild(); } secondLargest = parent; } else if (node.hasLeftChild()) { child = node.leftChild(); while (child.hasRightChild()) { child = child.rightChild(); } secondLargest = child; } } return secondLargest; }

Roshan on

0

For kth smallest, descend the left subtree first. class Node: def __init__(self, value, left=None, right=None): self.value = value self.left = left self.right = right def findKthLargest(root, k): global count if root is None: return findKthLargest(root.right, k) count += 1 if count == k: print root.value return findKthLargest(root.left, k) count = 0 r = Node(10, Node(5, Node(2), Node(7)), Node(30, Node(22), Node(32))) findKthLargest(r, 3)

General case for kth largest in python on

0

// solution in java // main routine Node findSecondMax(Node root) { if(root == null || (root.left == null && root.right == null) return null; else { Node max = findMax(root); return (max.parent == null) ? findMax(max.left) : max.parent; } } //helper routine, recursive implementation.... can also be done non-recursively Node findMax(Node root) { return (root.right == null) ? root : findMax(root.right); }

ytn01 on

0

Find the largest number in the binary tree and delete it. And again find the largest number. Short and fast.

Anonymous on

0

Reverse in-order traversal of the BST, keeping a count of # of visited nodes. This methods works great to return the kth largest element in a BST.

Anonymous on

0

mindpower's solution looks right

Anonymous on

3

In-order traverse the tree. The second last element in the array in the answer.

test on

0

Soln by "mindpower" works. Thank you. I am trying to solve a similar problem Find the 2nd nearest high(in in-order traversal) value for a given node Eg: Given nums: 12 7 14 3, construct a BST. If the given value is: 7 then we should return 14 (in the sort order: 3, 7, 12, 14) if the given value is: 3 then we should return 12 (in the sort order: 3, 7, 12, 14)

vesadi on

0

Generic solution in C# for any k. Notice that this example can be easily changed to find the k-th smallest node by doing a depth-first recursion on root.Left first, and then a tail recursion on root.Right. public Node GetKthLargest(int k) { return GetKthLargest(ref k, this.Root); } Node GetKthLargest(ref int k, Node root) { if (root == null || k < 1) return null; var node = GetKthLargest(ref k, root.Right); if (node != null) return node; if (--k == 0) return root; return GetKthLargest(ref k, root.Left); }

Erhhung on

0

In Python: def find_second_largest_bst_element(root, parent=None): if parent is None: # BST root if root.right is None: # no right subtree if root.left is not None: # if a left subtree exists... return root.left else: # root is the only element of the BST return False else: if root.right is None: # right-most element if root.left is not None: # left subtree exists return root.left else: # leaf return parent else: # check right subtree find_second_largest_bst_element(root.right, root) find_second_largest_bst_element(root)

d_t on

2

recursion is not needed. SecondLargest(Node root, Node secondLarge) { if(root.right==null) return root.left; Node secondLargest = root; while(secondLargest.right.right==null) secondLargest=secondLargest.right; return secondLargest; }

Techie on

1

if (root == null || (!root.hasRightChild() ) { return null;} else return findSecondGreatest(root, root.getValue()); value findSecondGreatest(Node curr, value oldValue) { if(curr.hasRightChild()) { return (findSecondGreatest( curr.getRightChild(), curr.value)); } else return oldValue; }

Anon on

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