**Difficulty:** Medium, **Asked-in:** Amazon, Google

**Key takeaway:** an excellent problem to learn problem-solving using **inorder** traversal and BST **augmentation** (storing extra information inside BST nodes for solving a problem).

Given the root of a binary search tree and an integer k, write a program to return the kth largest value among all the node values in the given BST.

- If the tree is empty, return INT_MIN.
- Assume that all values in the BST nodes are unique.

```
Input:
12
/ \
7 17
/ \ /
3 9 14
If k = 2, then 14 is the 2nd largest element.
If k = 4, then 9 is the 4th largest element.
If k = 5, then 7 is the 5th largest element.
```

**Important note:** before moving on to the solutions, we recommend trying this problem on paper for atleast 15 or 30 minutes. Enjoy problem-solving!

- Brute force approach using inorder traversal and extra space
- Recursive in-place solution: reverse in-order traversal
- Iterative in-place solution: reverse in-order traversal using stack
- Efficient solution using BST augmentation

As we know, a binary search tree is a sorted version of the binary tree where each node value is greater than all the node values in the left subtree and greater than all the node values in the right sub-tree. So if we perform an in-order traversal, it will produce elements in increasing or sorted order. Think!

So a basic idea would be:

- Traverse BST using inorder traversal and store elements in an extra array.
- Now n array elements are arranged in increasing order, and we can easily find the kth largest element by accessing the element at
**(n - k)th**index from the start. Note: as we know that all elements are unique. So the largest element will be present at (n -1)th index of the array.

```
// Inorder traversal of the BST to get the elements in sorted order
void inorder(BSTNode* root, vector<int>& SortedArray)
{
if (root == NULL)
return;
inorder(root->left, SortedArray);
SortedArray.push_back(root->data);
inorder(root->right, SortedArray);
}
// Function to find the kth largest element in the BST
int kthLargestBST(BSTNode* root, int k)
{
if (root == NULL)
return INT_MIN;
vector<int> SortedArray;
inorder(root, SortedArray);
int n = SortedArray.size();
return SortedArray[n - k];
}
```

```
# Inorder traversal of the BST to get the elements in sorted order
def inorder(root, SortedArray):
if root is None:
return
inorder(root.left, SortedArray)
SortedArray.append(root.data)
inorder(root.right, SortedArray)
# Function to find the kth largest element in the BST
def kthLargestBST(root, k):
if root is None:
return -sys.maxsize
SortedArray = []
inorder(root, SortedArray)
n = len(SortedArray)
return SortedArray[n - k]
```

Time complexity = Time complexity of inorder traversal to store elements in a sorted order+ Accessing (n - k)th element from the sorted list = O(n) + O(1) = **O(n)**. Space complexity = **O(n)**, we are using an extra array list of size n.

The critical question is: can we solve the problem without using extra space? can we track the kth largest element using the in-order traversal? Think!

**In-order traversal:**left subtree -> root -> right subtree. We first access the leftmost node in in-order traversal which stores the smallest value of BST. In other words, it explores all the nodes in increasing order.**Reverse in-order traversal:**right subtree -> root -> left subtree. If we perform reverse inorder traversal, it will first explore the rightmost node that stores the largest element of BST. In other words, it explores all the nodes in decreasing order. Think!

So one idea would be: while doing the reverse inorder traversal, we keep decreasing the value of k by one after accessing each node. When the k becomes equal to 0, we stop the traversal because the current node is our kth largest. So we return the value stored in the current node. This way, one could speed up the solution because there is no need to do complete inorder traversal, and one could stop after finding the kth largest element.

```
// Function to find the kth largest element in the BST
int kthLargestBST(BSTNode* root, int& k)
{
if (root == NULL)
return INT_MIN;
int right = kthLargestBST(root->right, k);
if (k == 0)
return right;
k = k - 1;
if (k == 0)
return root->data;
return kthLargestBST(root->left, k);
}
```

```
# Function to find the kth largest element in the BST
def kthLargestBST(root, k):
if root is None:
return -sys.maxsize
right = kthLargestBST(root.right, k)
if k[0] == 0:
return right
k[0] = k[0] - 1
if k[0] == 0:
return root.data
return kthLargestBST(root.left, k)
```

At each step of traversal, we are accessing nodes in decreasing order and decrementing the value of k by 1. So in general, our traversal will return value after the k number of steps. So time complexity = k * time complexity of each step = k * O(1) = **O(k).**

- In the worst case, k = n, so the worst-case time complexity = O(n)
- In the worst case, k = 1, so the best-case time complexity = O(1)

Space complexity = **O(h)** for recursion call stack, where h is the height of the tree.

This idea is similar to the above approach, but instead of using recursive inorder traversal in reverse order, we are using iterative reverse traversal with the help of a stack. The idea here is: when we pop node from the stack, we decrement the value of k by 1 and check k is equal to 0 or not. If yes, we stop and return the value stored in the current node.

you can explore this blog: iterative tree traversal using stack.

```
int kthLargestBST(BSTNode* root, int k)
{
if (root == NULL)
return INT_MIN;
stack<BSTNode*> BSTStack;
BSTNode* curr = root;
while (!BSTStack.empty() || curr != NULL)
{
if (curr != NULL)
{
BSTStack.push(curr);
curr = curr->right;
}
else
{
curr = BSTStack.top();
BSTStack.pop();
k = k - 1;
if (k == 0)
return curr->data;
curr = curr->left;
}
}
}
```

```
def kthLargestBST(root, k):
if root is None:
return -sys.maxsize
stack = []
curr = root
while stack or curr is not None:
if curr is not None:
stack.append(curr)
curr = curr.right
else:
curr = stack.pop()
k = k - 1
if k == 0:
return curr.data
curr = curr.left
```

As mentioned in the analysis of the previous approach, the iterative traversal will also access nodes in decreasing order by decrementing the value of k. So iteration will stop after k number of steps. Time complexity = k * O(1) = O(k).

- The worst-case time complexity = O(n)
- The best-case time complexity = O(1)

Space complexity = O(h) for recursion call stack, where h is the height of the tree.

The critical question is: what if the BST is often modified using insert or delete operations, and we need to find the kth largest element frequently? Can we optimize the time complexity? Let's think!

We know from the BST property that the value stored in all the nodes in the right sub-tree of any node is larger than the value stored in that node. So for the given root in the BST, if we know the count of elements in its right sub-tree, we can easily define the rank or order of the root node. Suppose there are 5 elements in the right subtree, then the root element would be the 6th largest element in the tree.

**Step 1:** We define a new structure of the BST node where we use an extra parameter **rightCount** to store the node count in the right sub-tree.

```
struct BSTNode
{
int data;
BSTNode* left;
BSTNode* right;
int rightCount;
BSTNode(int value)
{
data = value;
left = right = NULL;
rightCount = 0;
}
};
```

**Step 2:** now we build a BST of the given input by inserting each element one by one. During the insertion, we can keep track of a number of elements in the right subtree and update the rightCount for each node.

Implementation code c++

```
BSTNode* insertBST(BSTNode* root, int value)
{
if (root == NULL)
return new BSTNode(value);
if (value > root->data)
{
root->right = insertBST(root->right, value);
root->rightCount = root->rightCount + 1;
}
else if (value < root->data)
root->left = insertBST(root->left, value);
return root;
}
```

Implementation code Python

```
def insertBST(root, value):
if root is None:
return BSTNode(value)
if value > root.data:
root.right = insertBST(root.right, value)
root.rightCount = root.rightCount + 1
elif value < root.data:
root.left = insertBST(root.left, value)
return root
```

**Step 3:** Now we define a function **kthLargestBST(root, k)** to return the kth largest element.

We start from the root node by comparing k with the order or rank of the root node. What would be the rank of the root node? The answer is simple: If there are root-> rightCount nodes in the right subtree, then the root is (root-> rightCount + 1)th largest element in the tree. Think!

**If (k = root-> rightCount + 1):**the root->data is the required kth maximum element and we return this value as an output.**If (k < root-> rightCount + 1):**the kth largest would be present in the right sub-tree because the rank of the root is greater than the kth largest element. So kth largest element of the overall tree would also be the kth largest element of the right subtree. So we search the kth largest element recursively in the right subtree i.e. kthLargestBST(root->right, k).**If (k > root-> rightCount + 1):**the kth largest would be present in the left sub-tree because the rank of the root is smaller than the kth largest element. So we search the kth largest element recursively in the right subtree i.e. kthLargestBST(root->left, k - rightCount - 1). Note: Why are we passing**k - rightCount - 1**as the input parameter? The answer is simple: we are searching the kth largest element in the left subtree, and (rightCount + 1)number of elements are already greater than all the elements in the left subtree. So the rank of the kth largest element in the left subtree would be k - rightCount - 1. Think!

Implementation code C++

```
int kthLargestBST(BSTNode* root, int k)
{
if (root == NULL)
return INT_MIN;
if (k == root->rightCount + 1)
return root->data;
if (k < root->rightCount + 1)
return kthLargestBST(root->right, k);
else
return kthLargestBST(root->left, k - root->rightCount - 1);
}
```

Implementation code Python

```
def kthLargestBST(root, k):
if root is None:
return -sys.maxsize
if k == root.rightCount + 1:
return root.data
if k < root.rightCount + 1:
return kthLargestBST(root.right, k)
else:
return kthLargestBST(root.left, k - root.rightCount - 1)
```

Implementation code C++

```
int kthLargestBST(BSTNode* root, int k)
{
if (root == NULL)
return INT_MIN;
BSTNode* curr = root;
while (curr != NULL)
{
if (k == curr->rightCount + 1)
return curr->data;
else if (k < curr->rightCount + 1)
curr = curr->right;
else
{
k = k - curr->rightCount - 1;
curr = curr->left;
}
}
}
```

Implementation code Python

```
def kthLargestBST(root, k):
if root is None:
return -sys.maxsize
curr = root
while curr is not None:
if k == curr.rightCount + 1:
return curr.data
elif k < curr.rightCount + 1:
curr = curr.right
else:
k = k - curr.rightCount - 1
curr = curr.left
```

The time complexity of insert and delete operations into a BST is O(h), where h is the height of the binary tree. h is O(logn) for the balanced tree and O(n) for a skewed tree.

Once our BST is ready with the rightCount of each node, we are searching the kth largest element by going left or right at each step based on the comparison. In other, we reduce the input size by the size of the left or right subtree at each step. So in the worst case, we will be traversing the tree's height and doing an O(1) operation at each step. So time complexity of finding kth largest = O(h). Think!

Space complexity = storing rightCount inside each node = O(n). For recursive implementation, we also need to also consider the size of the recursion call stack which is O(h). But overall space complexity would be O(n) in both implementations. Think!

- How can we modify the above approaches to find Kth smallest in a BST?
- Design an algorithm to find the kth largest in a binary tree.
- What is the worst and best-case scenario of space complexity in the above approaches?
- Instead of reverse inorder traversal, can we solve it using in-order traversal?
- How can we solve this problem using Morris traversal?
- Can we think of solving this problem using max or minheap?
- What would be the time complexity of building BST of n nodes?
- If BST is balanced, the time complexity would be O(logn). So in the scenario of frequent insert or delete operations, BST may get imbalanced. How can we modify the above insert operation to ensure a balance BST? We should explore the idea of AVL or a red-black tree.

- Inorder traversal and extra space: Time = O(n), Space = O(n)
- Reverse in-order traversal (recursive): Time = O(k), Space = O(h)
- Reverse in-order traversal (iterative): Time = O(k), Space = O(h)
- Using BST augmentation: Time = O(h), Space = O(n)

- Kth smallest element in a BST
- Kth largest in an unsorted array
- Merge two BSTs with constant extra space
- Convert Sorted Array to Binary Search Tree

Please write in the message below if you find anything incorrect, or you want to share more insight, or you know some different approaches to solve this problem. Enjoy learning, Enjoy algorithms!

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