From 5a44b40b64918187b02a676a9739153e60eb1c30 Mon Sep 17 00:00:00 2001
From: gameloader <ggwqqo@163.com>
Date: Tue, 27 Feb 2024 11:18:19 +0800
Subject: [PATCH] add leetcode post

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 1 file changed, 73 insertions(+)
 create mode 100644 content/posts/leetcode.md

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+---
+title: leetcode everyday
+author: Logic
+date: 2024-02-27
+categories: [""]
+tags: []
+draft: false
+---
+
+## day 1 2024-02-27
+
+### 543 Diameter of Binary Tree
+
+Given the root of a binary tree, return the length of the diameter of the tree.
+
+The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.
+
+The length of a path between two nodes is represented by the number of edges between them.
+
+### 题解
+
+读题, 题目要求寻找二叉树中任意两节点之间的最大距离. 这时这个二叉树更像是一个图的性质而不是树, 即寻找图中任意两节点直接的最大距离. 考虑任意一点到另一点的距离等于其到另一点的父节点的距离减一, 则使用一个二维数组保存每个节点的两个属性:
+
+1. 以该节点为根节点且经过该节点的最大直径
+2. 以该节点为根节点的子树中叶子节点到该节点的最大距离
+
+属性1可以通过将该节点的两个子节点的属性2加和并加1来计算. 属性2取两个子节点属性2的最大值并加1来计算. 最后遍历数组求得数组中属性1的最大值即可. 有一点动态规划的思想.
+
+题目中并没有提供二叉树的节点总数, 则可以使用动态创建的方法, 在遍历过程中每遇到新节点就在二维数组中增加一项. 这里使用递归来对树进行后序遍历, 对空节点, 设置其属性2的值为-1, 这样保证叶子节点的属性2的值为0.
+
+解题时遇到一个小bug, 使用了一个全局切片(数组)来保存变量时, 第一个测试用例的数据会保留到测试第二个测试用例的过程中, 这大概是因为leetcode的测试是对每个用例直接调用给出的解题入口函数, 因此需要在解题函数中将使用的全局变量初始化一下, 将数组设置为空后问题得到解决.
+
+### 代码
+
+```go
+/**
+ * Definition for a binary tree node.
+ * type TreeNode struct {
+ *     Val int
+ *     Left *TreeNode
+ *     Right *TreeNode
+ * }
+ */
+
+var length [][]int
+
+func diameterOfBinaryTree(root *TreeNode) int {
+    length = nil
+    _ = postorder(root)
+    max := -1
+    for _, value := range length{
+        if value[0] > max{
+            max = value[0]
+        }
+    }
+    return max
+}
+
+func postorder(father *TreeNode) int {
+    if father != nil{
+        len1 := postorder(father.Left)
+        len2 := postorder(father.Right)
+        len := make([]int,2)
+        // find the max diameter pass through current node from the tree rooted current node
+        len[0] = len1 + len2 + 2
+        len[1] = max(len1, len2) + 1
+        length = append(length, len)
+        return len[1]
+    } else {
+        return -1
+    }
+}
+```