安橙的博客

环境准备 基础查询 约束与范式基础 索引 + EXPLAIN 事务与并发控制 附录 docker-componse.yml ...

二叉树的右视图 题目 思路 思路 1： 层序遍历，用队列做 BFS，最后出列的就是最右边。 ...

题目 思路 思路一. 中序遍历，取第 K 个 思路二. 维护每个子树的大小 ...

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附录 缓存系统实战 缓存系统实战 application.properties: ...

二叉树的层序遍历 题目 思路 put root into a queue queue is not empty 2.1. size = queue.size(), this is exactly how many nodes are in the current level. 2.1. repeat size times: * pop one node from queue * add its value to level * add left and right child Java 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 class Solution { public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> res = new ArrayList<>(); if (root == null) { return res; } Deque<TreeNode> q = new ArrayDeque<>(); q.offer(root); while(!q.isEmpty()) { int size = q.size(); List<Integer> level = new ArrayList<>(size); for (int i = 0; i < size; i ++) { TreeNode node = q.poll(); level.add(node.val); if (node.left != null) q.offer(node.left); if (node.right != null) q.offer(node.right); } res.add(level); } return res; } }

题目 Solution The longest path through that node equals: left subtree depth + right subtree depth. the longest path must pass through some node as a highest point. ...

将有序数组转换为二叉搜索树 题目 思路 A height-balance BST is achieved by always picking the middle element as the root. The left and right subtree sizes differ by at most 1. ...

验证二叉搜索树 题目 思路 root starts with (-inf, +inf) left child inherits (min, root.val) right child inherits (root.val, max) Java 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 class Solution { public boolean isValidBST(TreeNode root) { return dfs(root, Long.MIN_VALUE, Long.MAX_VALUE); } private boolean dfs(TreeNode node, long low, long high) { if (node == null) { return true; } long v = node.val; if (v <= low || v >= high) { return false; } return dfs(node.left, low, v) && dfs(node.right, v, high); } }