Closest Binary Search Tree Value II
題意：
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.
解題思路：
使用priority queue，因queue中頂端的值是queue中所有元素最小的值，表示與target差距最大，所以只要不斷的與queue頂端的元素作比較即可，若數目已達k，且頂端元素與target的差距比新的元素與target的差距還大的話，則把頂端元素pop掉並插入新的元素，程式碼如下：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private PriorityQueue<Integer> q;
private int count = 0;
public List<Integer> closestKValues(TreeNode root, double target, int k) {
q = new PriorityQueue<Integer>();
List<Integer> res = new ArrayList<>();
if (root == null || k == 0) {
return res;
}
inorderTraverse(root, target , k);
for (Integer val : q) {
res.add(val);
}
return res;
}
private void inorderTraverse(TreeNode root, double target, int k) {
if (root == null) {
return;
}
inorderTraverse(root.left, target, k);
if (count < k) {
q.offer(root.val);
} else {
if (Math.abs((double) root.val - target) < Math.abs((double)q.peek() - target)) {
q.poll();
q.offer(root.val);
}
}
count++;
inorderTraverse(root.right, target , k);
}
}
另外可維護兩個predecessors與sucessors的stack，接著再從兩個輸出最小的k個數，其程式碼如下：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
Stack<Integer> predecessors;
Stack<Integer> successors;
public List<Integer> closestKValues(TreeNode root, double target, int k) {
List<Integer> res = new ArrayList<>();
predecessors = new Stack<>();
successors = new Stack<>();
if (root == null || k == 0) {
return res;
}
getPredecessors(root, target);
getSuccessors(root, target);
for (int i = 0; i < k; i++) {
if (predecessors.isEmpty()) {
res.add(successors.pop());
} else if (successors.isEmpty()) {
res.add(predecessors.pop());
} else if (Math.abs((double)predecessors.peek() - target) < Math.abs((double)successors.peek() - target)) {
res.add(predecessors.pop());
} else {
res.add(successors.pop());
}
}
return res;
}
private void getPredecessors(TreeNode root, double target) {
if (root == null) {
return;
}
getPredecessors(root.left, target);
if (root.val > target) {
return;
}
predecessors.push(root.val);
getPredecessors(root.right, target);
}
private void getSuccessors(TreeNode root, double target) {
if (root == null) {
return;
}
getSuccessors(root.right, target);
if (root.val <= target) {
return;
}
successors.push(root.val);
getSuccessors(root.left, target);
}
}