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Commit 39ea331

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commit solution 111
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Lines changed: 44 additions & 4 deletions
Original file line numberDiff line numberDiff line change
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# [100. xxx](https://leetcode-cn.com/problems/recover-binary-search-tree)
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# [111. 二叉树的最小深度](https://leetcode-cn.com/problems/minimum-depth-of-binary-tree/description/)
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### 题目描述
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<p>给定一个二叉树,找出其最小深度。</p>
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<p>最小深度是从根节点到最近叶子节点的最短路径上的节点数量。</p>
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<p><strong>说明:</strong>&nbsp;叶子节点是指没有子节点的节点。</p>
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<p><strong>示例:</strong></p>
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<p>给定二叉树&nbsp;<code>[3,9,20,null,null,15,7]</code>,</p>
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<pre> 3
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/ \
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9 20
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/ \
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15 7</pre>
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<p>返回它的最小深度 &nbsp;2.</p>
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### 解题思路
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### 具体解法
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<!-- tabs:start -->
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1. BFS
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#### **Golang**
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```go
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func minDepth(root *TreeNode) int {
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if root == nil {
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return 0
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}
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count := 1
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queue := []*TreeNode{root}
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for len(queue) > 0 {
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l := len(queue)
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for i := 0; i < l; i++ {
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node := queue[i]
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if node.Left == nil && node.Right == nil {
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return count
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}
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if node.Left != nil {
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queue = append(queue, node.Left)
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}
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if node.Right != nil {
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queue = append(queue, node.Right)
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}
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}
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count++
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queue = queue[l:]
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}
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return count
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}
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```
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<!-- tabs:end -->
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package leetcode
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/*
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* @lc app=leetcode.cn id=111 lang=golang
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*
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* [111] 二叉树的最小深度
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*/
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// @lc code=start
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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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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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func minDepth(root *TreeNode) int {
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if root == nil {
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return 0
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}
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count := 1
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queue := []*TreeNode{root}
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for len(queue) > 0 {
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l := len(queue)
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for i := 0; i < l; i++ {
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node := queue[i]
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if node.Left == nil && node.Right == nil {
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return count
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}
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if node.Left != nil {
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queue = append(queue, node.Left)
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}
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if node.Right != nil {
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queue = append(queue, node.Right)
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}
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}
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count++
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queue = queue[l:]
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}
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return count
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}
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// @lc code=end

‎solution/200-299/0242.valid-anagram/solution_test.go

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Original file line numberDiff line numberDiff line change
@@ -4,14 +4,32 @@ import (
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"testing"
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)
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func TestHammingWeight(t *testing.T) {
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var num uint32
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var ret int
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func TestIsAnagram(t *testing.T) {
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var ss string
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var ts string
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var ret bool
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num = 00000000000000000000000000001011
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ret = 3
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ss = "dmomkifm"
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ts = "skmokj"
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ret = false
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if ret != hammingWeight(num) {
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if ret != isAnagram(ss, ts) {
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t.Fatalf("case fails %v\n", ret)
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}
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ss = "ddss"
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ts = "ssdd"
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ret = true
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if ret != isAnagram(ss, ts) {
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t.Fatalf("case fails %v\n", ret)
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}
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ss = "ddsse"
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ts = "ssddd"
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ret = false
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if ret != isAnagram(ss, ts) {
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t.Fatalf("case fails %v\n", ret)
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}
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}

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