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76 lines
2.8 KiB
Markdown
76 lines
2.8 KiB
Markdown
---
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title: leetcode everyday
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author: Logic
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date: 2024-02-27
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categories: [""]
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tags: []
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draft: false
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---
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## day 1 2024-02-27
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### 543 Diameter of Binary Tree
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Given the root of a binary tree, return the length of the diameter of the tree.
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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.
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The length of a path between two nodes is represented by the number of edges between them.
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### 题解
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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之间的最大距离. 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:
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1. 以该节点为根节点且经过该节点的最大直径
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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
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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.
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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.
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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.
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### 代码
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```go
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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var length [][]int
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func diameterOfBinaryTree(root *TreeNode) int {
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length = nil
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_ = postorder(root)
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max := -1
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for _, value := range length{
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if value[0] > max{
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max = value[0]
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}
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}
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return max
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}
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func postorder(father *TreeNode) int {
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if father != nil{
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len1 := postorder(father.Left)
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len2 := postorder(father.Right)
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len := make([]int,2)
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// find the max diameter pass through current node from the tree rooted current node
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len[0] = len1 + len2 + 2
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len[1] = max(len1, len2) + 1
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length = append(length, len)
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return len[1]
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} else {
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return -1
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}
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}
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```
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