Understanding the Problem
We are given a Binary Search Tree (BST), and we need to find the k
th largest element in it. A BST has the property that for any given node, the left subtree contains values less than the node, and the right subtree contains values greater than the node.
To solve this, we can utilize reverse in-order traversal (right → node → left), which visits nodes in descending order. This approach allows us to find the k
th largest element efficiently by counting how many nodes we’ve visited so far during the traversal.
Step-by-Step Solution with Example
Step 1: Choose an Example
Let’s consider the example BST: [5,3,6,2,4,null,null,1]
.
This represents the following BST structure:
5
/ 3 6
/ 2 4
/
1
We are asked to find the k = 3
rd largest element.
Step 2: Understand Reverse In-Order Traversal
In normal in-order traversal (Left → Node → Right), we get sorted ascending order. But here, we use reverse in-order (Right → Node → Left) to get descending order. As we visit each node, we keep a count.
Step 3: Traverse and Count
Start from the rightmost node (which is the largest):
- Visit 6 → count = 1
- Backtrack to 5 → count = 2
- Go to left child 3, then right child 4 → count = 3
Since count = 3 at node 4, we return 4 as the answer.
Step 4: Code Outline
We define a helper function to do reverse in-order traversal recursively, while maintaining a counter and storing the result once we reach the k
th element.
// Java-style pseudocode
int count = 0;
int result = -1;
void reverseInorder(TreeNode node, int k) {
if (node == null || count >= k) return;
reverseInorder(node.right, k);
count++;
if (count == k) {
result = node.val;
return;
}
reverseInorder(node.left, k);
}
Edge Cases
Case 1: k = 1 (Largest Element)
When k = 1
, we want the largest element. Since the rightmost node in a BST is the largest, we can directly return the value of the rightmost node if we reach it first during traversal.
Case 2: Small BST
For a small BST like [2,1,3]
, reverse traversal visits nodes in the order [3,2,1]. If k = 2
, the answer is 2.
Case 3: Single Node Tree
If the tree has just one node, say [5]
, then the only valid k
is 1. The 1st largest element is the node itself: 5.
Case 4: Empty Tree
If the tree is empty, there are no elements to return. Any k
in this scenario is invalid, and we must return an error or handle it gracefully (e.g., return -1 or throw an exception).
Case 5: k Greater Than Number of Nodes
If k
is larger than the total number of nodes, the traversal will finish without finding the kth largest. We should check node count beforehand or handle this as an invalid input.
Finally
This problem teaches us how to apply a traversal technique that respects the properties of BSTs to solve problems efficiently. By understanding reverse in-order traversal and maintaining a count, we can find the k
th largest element in O(H + k)
time, where H
is the height of the tree. Always remember to consider edge cases like empty trees, invalid k
values, and small tree structures during implementation.
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