題目

https://leetcode.com/problems/kth-smallest-element-in-a-bst/

題解

方法1：中序遍歷+剪枝

import java.util.ArrayList; import java.util.List;class TreeNode {int val;TreeNode left;TreeNode right;TreeNode() {}TreeNode(int val) { this.val = val; }TreeNode(int val, TreeNode left, TreeNode right) {this.val = val;this.left = left;this.right = right;} }class Solution {List<Integer> list;int k;public int kthSmallest(TreeNode root, int k) {list = new ArrayList<>();this.k = k;inOrder(root);return list.get(k - 1);}public void inOrder(TreeNode node) {if (node == null) return;inOrder(node.left);list.add(node.val);if (list.size() >= k) return; // 剪枝，及時停止，不需要走完整棵樹inOrder(node.right);} }

方法2：Iterative Inorder Traversal

參考：https://leetcode.com/problems/kth-smallest-element-in-a-bst/solution/
The above recursion could be converted into iteration, with the help of stack. This way one could speed up the solution because there is no need to build the entire inorder traversal, and one could stop after the kth element.

class Solution {public int kthSmallest(TreeNode root, int k) {LinkedList<TreeNode> stack = new LinkedList<TreeNode>();while (true) {while (root != null) {stack.add(root);root = root.left;}root = stack.removeLast();if (--k == 0) return root.val;root = root.right;}} }

總結

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