package com.fishercoder.solutions;

import com.fishercoder.common.classes.TreeNode;

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

/**
 * 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.

 */
public class _272 {

	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;
	}

	// inorder traversal
	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);
	}

}

package com.fishercoder.solutions;

import com.fishercoder.common.classes.TreeNode;

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

/**
 * 272. Closest Binary Search Tree Value II
 *
 * 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.

 */
public class _272 {

    public static class Solution1 {
        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;
        }

        // inorder traversal
        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);
        }
    }
}

refactor 272