一、235. 二叉搜索树的最近公共祖先
题目链接:https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-search-tree/
文章讲解:https://programmercarl.com/0235.%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E7%9A%84%E6%9C%80%E8%BF%91%E5%85%AC%E5%85%B1%E7%A5%96%E5%85%88.html
视频讲解:https://www.bilibili.com/video/BV1Zt4y1F7ww?share_source=copy_web
1.1 初见思路
- 如何利用二叉搜索树的特性来实现?
1.2 具体实现
class Solution {public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {if (root.val > p.val && root.val > q.val)return lowestCommonAncestor(root.left, p, q);if (root.val < p.val && root.val < q.val)return lowestCommonAncestor(root.right, p, q);return root;}
}
1.3 重难点
- 如何分析出第一次遇到 cur节点是数值在[q, p]区间中,此时可以说明 q 和 p 一定分别存在于的左子树,和右子树中
二、 701.二叉搜索树中的插入操作
题目链接:https://leetcode.cn/problems/insert-into-a-binary-search-tree/
文章讲解:https://programmercarl.com/0701.%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E4%B8%AD%E7%9A%84%E6%8F%92%E5%85%A5%E6%93%8D%E4%BD%9C.html
视频讲解:https://www.bilibili.com/video/BV1Et4y1c78Y
2.1 初见思路
- 可以不需要重构
- 找到空节点,就可以插入
2.2 具体实现
class Solution {public TreeNode insertIntoBST(TreeNode root, int val) {if (root == null)return new TreeNode(val);TreeNode newRoot = root;TreeNode pre = root;while (root != null) {pre = root;if (root.val > val) {root = root.left;} else if (root.val < val) {root = root.right;}}if (pre.val > val) {pre.left = new TreeNode(val);} else {pre.right = new TreeNode(val);}return newRoot;}
}
2.3 重难点
三、 450.删除二叉搜索树中的节点
题目链接:https://leetcode.cn/problems/delete-node-in-a-bst/
文章讲解:https://programmercarl.com/0450.%E5%88%A0%E9%99%A4%E4%BA%8C%E5%8F%89%E6%90%9C%E7%B4%A2%E6%A0%91%E4%B8%AD%E7%9A%84%E8%8A%82%E7%82%B9.html
视频讲解:https://www.bilibili.com/video/BV1tP41177us
3.1 初见思路
3.2 具体实现
class Solution {public TreeNode deleteNode(TreeNode root, int key) {if (root == null)return root;if (root.val == key) {if (root.left == null) {return root.right;} else if (root.right == null) {return root.left;} else {TreeNode cur = root.right;while (cur.left != null) {cur = cur.left;}cur.left = root.left;root = root.right;return root;}}if (root.val > key)root.left = deleteNode(root.left, key);if (root.val < key)root.right = deleteNode(root.right, key);return root;}
}