leetcode 272: Closest Binary Search Tree Value II

Closest Binary Search Tree Value II

Total Accepted: 984 Total Submissions: 3704 Difficulty: Hard

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:

1. Consider implement these two helper functions:
   1. getPredecessor(N), which returns the next smaller node to N.
   2. getSuccessor(N), which returns the next larger node to N.
2. Try to assume that each node has a parent pointer, it makes the problem much easier.
3. Without parent pointer we just need to keep track of the path from the root to the current node using a stack.
4. You would need two stacks to track the path in finding predecessor and successor node separately.

[思路]

prefix traverse. 同时维护一个大小为k的 max heap. 注意根据bst的性质,在diff 大于 maxHeap时, 可以只遍历一边的子树.

[CODE]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    
    public List<Integer> closestKValues(TreeNode root, double target, int k) {
        PriorityQueue<Double> maxHeap = new PriorityQueue<Double>(k, new Comparator<Double>() { 
            @Override
            public int compare(Double x, Double y) {
                return (int)(y-x);
            }
        });
        Set<Integer> set = new HashSet<Integer>();
        
        rec(root, target, k, maxHeap, set);
        
        return new ArrayList<Integer>(set);
    }
    
    private void rec(TreeNode root, double target, int k, PriorityQueue<Double> maxHeap, Set<Integer> set) {
        if(root==null) return;
        double diff = Math.abs(root.val-target);
        if(maxHeap.size()<k) {
            maxHeap.offer(diff);
            set.add(root.val);
        } else if( diff < maxHeap.peek() ) {
            double x = maxHeap.poll();
            if(! set.remove((int)(target+x))) set.remove((int)(target-x));
            maxHeap.offer(diff);
            set.add(root.val);
        } else {
            if(root.val > target) rec(root.left, target, k, maxHeap,set);
            else rec(root.right, target, k, maxHeap, set);
            return;
        }
        rec(root.left, target, k, maxHeap, set);
        rec(root.right, target, k, maxHeap, set);
    }
}