Cloest Binary Search Tree Value II

最新推荐文章于 2022-11-10 13:30:08 发布

转载最新推荐文章于 2022-11-10 13:30:08 发布 · 62 阅读

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原文链接：http://www.cnblogs.com/shuashuashua/p/5723603.html

本文介绍了一种高效的算法，用于在二叉搜索树中找到与目标值最接近的K个整数值。该算法通过初始化两个栈分别追踪前驱节点和后继节点，然后比较两栈顶节点的值与目标值的距离来确定添加哪个值到结果列表中。

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 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<Integer> closestKValues(TreeNode root, double target, int k) {
12         List<Integer> result = new ArrayList<>();
13         if (k == 0 || root == null) {
14             return result;
15         }
16         
17         Stack<TreeNode> successor = new Stack<>();
18         Stack<TreeNode> precessor = new Stack<>();
19         
20         initialStack(successor, root, true, target);
21         initialStack(precessor, root, false, target);
22         
23         if(!successor.isEmpty() && !precessor.isEmpty() && successor.peek().val == precessor.peek().val) {
24             getNext(precessor, false);
25         }
26         
27         while (result.size() < k) {
28             if (successor.isEmpty()) {
29                 result.add(getNext(precessor, false));
30             } else if (precessor.isEmpty()) {
31                 result.add(getNext(successor, true));
32             } else {
33                 boolean sclose = Math.abs((double) successor.peek().val - target) < Math.abs((double)precessor.peek().val - target);
34                 if (sclose) {
35                     result.add(getNext(successor, true));
36                 } else {
37                     result.add(getNext(precessor, false));
38                 }
39             }
40         }
41         return result;
42     }
43     
44     private void initialStack(Stack<TreeNode> stack, TreeNode root, boolean successor, double target) {
45         while (root != null) {
46             if (root.val == target) {
47                 stack.push(root);
48                 break;
49             } else if (root.val > target == successor) {
50                 stack.push(root);
51                 root = successor ? root.left : root.right;
52             } else {
53                 root = successor ? root.right : root.left;
54             }
55         }
56     }
57     
58     private int getNext(Stack<TreeNode> stack, boolean successor) {
59         TreeNode current = stack.pop();
60         int result = current.val;
61         current = successor ? current.right : current.left;
62         while (current != null) {
63             stack.push(current);
64             current = successor ? current.left : current.right;
65         }
66         return result;
67     }
68 }

转载于:https://www.cnblogs.com/shuashuashua/p/5723603.html