Closest Binary Search Tree Value II
Description
Given a non-empty binary search tree and a target value, find k values in the BST that are closest to the target.
Note: Given target value is a floating point. You may assume k is always valid, that is: k ≤ total nodes. You are guaranteed to have only one unique set of k values in the BST that are closest to the target. Follow up: Assume that the BST is balanced, could you solve it in less than O(n) runtime (where n = total nodes)?
Hint
Consider implement these two helper functions: getPredecessor(N), which returns the next smaller node to N. getSuccessor(N), which returns the next larger node to N. Try to assume that each node has a parent pointer, it makes the problem much easier. Without parent pointer we just need to keep track of the path from the root to the current node using a stack. You would need two stacks to track the path in finding predecessor and successor node separately.
Train of Thought
The idea is to compare the predecessors and successors of the closest node to the target, we can use two stacks to track the predecessors and successors, then like what we do in merge sort, we compare and pick the closest one to the target and put it to the result list.
As we know, inorder traversal gives us sorted predecessors, whereas reverse-inorder traversal gives us sorted successors.
We can use iterative inorder traversal rather than recursion, but to keep the code clean, here is the recursion version.
Code
public List<Integer> closestKValues(TreeNode root, double target, int k) {
List<Integer> res = new ArrayList<>();
Stack<Integer> s1 = new Stack<>(); // predecessors
Stack<Integer> s2 = new Stack<>(); // successors
inorder(root, target, false, s1);
inorder(root, target, true, s2);
while (k-- > 0) {
if (s1.isEmpty())
res.add(s2.pop());
else if (s2.isEmpty())
res.add(s1.pop());
else if (Math.abs(s1.peek() - target) < Math.abs(s2.peek() - target))
res.add(s1.pop());
else
res.add(s2.pop());
}
return res;
}
public void inorder(TreeNode root, double target, boolean reverse, Stack<Integer> stack) {
if (root == null) return;
inorder(reverse ? root.right : root.left, target, reverse, stack);
// early terminate, no need to traverse the whole tree
if ((reverse && root.val <= target) || (!reverse && root.val > target)) return;
// track the value of current node
stack.push(root.val);
inorder(reverse ? root.left : root.right, target, reverse, stack);
}