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.
Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?
Hint:
- Try to utilize the property of a BST.
- What if you could modify the BST node's structure?
- The optimal runtime complexity is O(height of BST).
Similar: Binary Tree Inorder Traversal
1 Solution 1. Binary Search
2 Time Complexity: T(n) = n/2 + T(n/2) = O(n)
3 /**
4 * Definition for a binary tree node.
5 * public class TreeNode {
6 * int val;
7 * TreeNode left;
8 * TreeNode right;
9 * TreeNode(int x) { val = x; }
10 * }
11 */
12 public class Solution {
13 public int kthSmallest(TreeNode root, int k) {
14 int count = countNodes(root.left);
15 if (k <= count) {
16 return kthSmallest(root.left, k);
17 } else if (k > count + 1) {
18 return kthSmallest(root.right, k-1-count);
19 }
20 return root.val;
21 }
22
23 public int countNodes(TreeNode n) {
24 if (n == null) return 0;
25 return 1 + countNodes(n.left) + countNodes(n.right);
26 }
27 }
28
29 Solution 2. Modify the structure of BST
30 Time Complexity: O(lg n)
31 /**
32 * Definition for a binary tree node.
33 * public class TreeNode {
34 * int val;
35 * TreeNode left;
36 * TreeNode right;
37 * TreeNode(int x) { val = x; }
38 * }
39 */
40 public class Solution {
41 public int kthSmallest(TreeNode root, int k) {
42 TreeWithCount twcRoot = buildTWC(root);
43 return findKth(twcRoot, k);
44 }
45
46 private int findKth(TreeWithCount root, int k) {
47 if (k <= 0 || k > root.count) return -1;
48 if (root.left != null) {
49 if (root.left.count == k-1) return root.val;
50 if (k <= root.left.count) return findKth(root.left, k);
51 return findKth(root.right, k-1-root.left.count);
52 } else {
53 if (k == 1) return root.val;
54 return findKth(root.right, k-1);
55 }
56 }
57
58 private TreeWithCount buildTWC(TreeNode root) {
59 if (root == null) return null;
60
61 TreeWithCount twcRoot = new TreeWithCount(root.val);
62 twcRoot.left = buildTWC(root.left);
63 twcRoot.right = buildTWC(root.right);
64 if (twcRoot.left != null) twcRoot.count += twcRoot.left.count;
65 if (twcRoot.right != null) twcRoot.count += twcRoot.right.count;
66 return twcRoot;
67 }
68
69 class TreeWithCount {
70 int val;
71 int count;
72 TreeWithCount left;
73 TreeWithCount right;
74 public TreeWithCount (int val) {
75 this.val = val;
76 this.count = 1;
77 }
78 }
79 }
来源：https://www.cnblogs.com/joycelee/p/5344428.html