Kth Smallest Element in a BST

最新推荐文章于 2022-04-14 14:33:13 发布

转载最新推荐文章于 2022-04-14 14:33:13 发布 · 63 阅读

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原文链接：http://www.cnblogs.com/FLAGyuri/p/5664259.html

本文介绍三种方法寻找二叉搜索树中第K小的元素：非递归中序遍历、递归中序遍历及通过统计左右子树节点数来确定目标元素位置。这些方法利用了二叉搜索树的特性。

Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.

Note:
You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

采用中序遍历的方法，找到第K小的数。考察理解BST的特性。

非递归中序遍历：

 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 int kthSmallest(TreeNode root, int k) {
12         Stack<TreeNode> stack = new Stack<>();
13         TreeNode curt = root;
14         while (!stack.empty() || curt != null) {
15             while (curt != null) {
16                 stack.push(curt);
17                 curt = curt.left;
18             }
19             curt = stack.pop();
20             --k;
21             if (k == 0) {
22                 return curt.val;
23             }
24             curt = curt.right;
25         }
26         return -1;
27     }
28 }

中序遍历递归：

 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     private int result = 0;
12     private int count = 0;
13     public int kthSmallest(TreeNode root, int k) {
14         count = k;
15         helper(root);
16         return result;
17     }
18     
19     private void helper(TreeNode root) {
20         if (root == null) {
21             return;
22         }
23         if (root.left != null) {
24             helper(root.left);
25         }
26         --count;
27         if (count == 0) {
28             result = root.val;
29             return;
30         }
31         if (root.right != null) {
32             helper(root.right);
33         }
34     }
35 }

分别统计左子树的节点数和右子树的节点数，根据左右子树的节点数与K的大小关系，知道第K小的数在左子树还是右子树。

 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 int kthSmallest(TreeNode root, int k) {
12         int count = countNodes(root.left);
13         if (k <= count) {
14             return kthSmallest(root.left, k);
15         } else if (k > count + 1) {
16             return kthSmallest(root.right, k - count - 1);
17         } else {
18                         //count + 1 == k
19             return root.val;
20         }
21     }
22 
23     private int countNodes(TreeNode root){
24         if(root == null) {
25             return 0;
26         }
27         return 1 + countNodes(root.left) + countNodes(root.right);
28     }
29 }