[LeetCode#272] Closest Binary Search Tree Value II

Problem:

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)?

Analysis:

This problem is not hard. But the description of the problem is really quite misleading. It strongly hints you to use the characteristics of balance tree. I fail to achieve that. But there is also a very innovative and efficient way to solve this problem. 

Ask: Where are closest nodes of a give number ? 
The predecessors and successors of a given number are the best candidates for you to choice, based on how many closest nodes you want.
And we already know through a inorder traversal we shoud easily get the tree in sorted form. It's really easy for us to find the closestKValues in the sorted form. 
1 2 3 5 6 [target: 7] 8 10 23 25

But for this problem, we could solve it in more elegant way. Use the idea we have used in merging k sorted list. 
Predecessors list: 6 5 3 2 1
Successors: 8 10 23 25

Basic knowledge enhancement
Through ordinary preorder traversal (scan left subtree first), we could get the binary search tree in the ascending order.
1 2 3 5 6 8 10 23 25
Through reverse preorder traversal (scan right subtree first), we could get the binary search tree in the descending order. 
25 23 10 8 6 5 3 2 1

Step 1: get the predecessors list : 6 5 3 2 1 (descending order), through a stack. 
------------------------------------------------------------------------
preOrderTraversal(is_reverse ? root.left, target, stack, is_reverse);
...
stack.push(root.val)
preOrderTraversal(is_reverse ? root.right, target, stack, is_reverse);
------------------------------------------------------------------------


Step 2: get the successors list : 8 10 23 25 (ascending order), through a stack.
------------------------------------------------------------------------
preOrderTraversal(is_reverse ? root.right, target, stack, is_reverse);
...
stack.push(root.val)
preOrderTraversal(is_reverse ? root.left, target, stack, is_reverse);
------------------------------------------------------------------------


A skill in getting "6 5 3 2 1" rather than "25 23 10 8 6 5 3 2 1".
if ((!is_reverse && root.val > target))
    return;
//Great during the traversal process, we stop when we reach a node larger than target.


A skill in getting "8 10 23 25" rahter than "25 23 10 8 6 5 3 2 1"
if (is_reverse && root.val <= target)
    return;
Note: we use the same termination skill at here.


Another skill to be careful (note when there is a stack was used up)
if (pre.isEmpty()) {
    ret.add(suc.pop());
} else if (suc.isEmpty()) {
    ret.add(pre.pop());
}
You must detect and handle above cases. 

Solution:

public class Solution {
    public List closestKValues(TreeNode root, double target, int k) {
        List ret = new ArrayList ();
        Stack pre = new Stack ();
        Stack suc = new Stack ();
        preOrderTraversal(root, target, pre, false);
        preOrderTraversal(root, target, suc, true);
        int count = 0;
        while (count < k) {
            if (pre.isEmpty()) {
                ret.add(suc.pop());
            } else if (suc.isEmpty()) {
                ret.add(pre.pop());
            } else if (Math.abs(target - pre.peek()) < Math.abs(target - suc.peek())) {
                ret.add(pre.pop());
            } else {
                ret.add(suc.pop());
            }
            count++;
        }
        return ret;
    }
    
    
    private void preOrderTraversal(TreeNode root, double target, Stack stack, boolean is_reverse) {
        if (root == null)
            return;
        preOrderTraversal(is_reverse ? root.right : root.left, target, stack, is_reverse);
        if ((is_reverse && root.val <= target) || (!is_reverse && root.val > target))
            return;
        stack.push(root.val);
        preOrderTraversal(is_reverse ? root.left : root.right, target, stack, is_reverse);
    }
}

 

转载于:https://www.cnblogs.com/airwindow/p/4823030.html