Java算法 二叉树入门 力扣简单题相同的树 翻转二叉树 判断对称二叉树 递归求二叉树的层数
目录
模版
先序遍历
中序遍历
后序遍历
力扣原题 相同的二叉树
力扣原题 翻转二叉树
遍历树的层数
题目
静态变量
核心逻辑
模版
// 二叉树public static class Node{public int value;public Node left;public Node right;public Node(int v) {value=v;}}先序遍历
根节点 左孩子节点 右孩子节点
中序遍历
左孩子节点 根节点 右孩子节点
后序遍历
左孩子节点 右孩子节点 根节点
力扣原题 相同的二叉树
100. 相同的树 - 力扣（LeetCode）
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution {public boolean isSameTree(TreeNode p, TreeNode q) {if(p==null && q!=null){return false;}if(p!=null && q==null){return false;}if(p==null && q==null){return true;}// 都不为空return p.val==q.val && isSameTree(p.left,q.left) && isSameTree(p.right,q.right);}
}力扣原题 翻转二叉树
/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode() {}* TreeNode(int val) { this.val = val; }* TreeNode(int val, TreeNode left, TreeNode right) {* this.val = val;* this.left = left;* this.right = right;* }* }*/
class Solution {public TreeNode flipTree(TreeNode root) {if(root == null) return null;TreeNode tmp = root.left;root.left = flipTree(root.right);root.right = flipTree(tmp);return root;}
}遍历树的层数
题目
静态变量
核心逻辑
import java.util.*;
import java.math.*;
import java.time.*;
import java.io.*;
import java.util.regex.*;// Eclipse IDE 2020 08
// OpenJDK 1.8
// 下方网页是个人主页
// http://gczdy.cn/*.-'''-.
_______ _______ ' _ \
\ ___ `'. \ ___ `'. / /` '. \ ' |--.\ \ ' |--.\ \ . | \ ' | | \ ' | | \ ' | ' | ' | | | '| | | ' \ \ / / | | | || | | | _ _`. ` ..' / | | ' .'| | ' .'| ' / | '-...-'` | |___.' /' | |___.' /'.' | .' |
/_______.'/ /_______.'/ / | / |
\_______|/ \_______|/ | `'. | ' .'| '/ `-' `--'
*//*** * @Author Dduo* @Date 2025-01-10*/
public class Main {// 普通流
// static Scanner sc = new Scanner(System.in);// 数据流快读模板(类似于C++的IOS)static Read sc=new Read();// 时间类 用来测试运行时间static Instant START=Instant.now();static long MOD = (long) (1e9 + 7);static int[] dx = {0, 0, 1, -1, 1, -1, 1, -1};static int[] dy = {1, -1, 0, 0, -1, -1, 1, 1};private static final int[] DIRECTIONS = {-1, 0, 1, 0, -1};/**** @param args* @return* @throws Exception */public static void main(String[] args) throws Exception {int t = 1;
// t = sc.nextInt();// 预处理preconditioning();while (t-- > 0) {solve();}// sc.close();
// dduoln("运行时间 : "+Duration.between(START,Instant.now()).toMillis()+"ms");return;}// 输出流 static <T> void dduoln(T t) {System.out.println(t);}static <T> void dduo(T t) {System.out.print(t);}// 预处理static void preconditioning() {}// 数据结构模板 二叉树 by Dduostatic class Node{public int value;public Node left;public Node right;public Node() {}public Node(int v) {value=v;}}// 静态变量static Node[] a = new Node[1000010];static int cnt=0;// 核心代码逻辑static void solve() throws Exception { // 构造二叉树int n=sc.nextInt();for(int i=1;i<=n;i++) {a[i]=new Node(i);int l=sc.nextInt();int r=sc.nextInt();if (l != 0) {a[i].left = new Node(l);}if (r != 0) {a[i].right = new Node(r);}}dfs(a[1],1);dduoln(cnt);}static void dfs(Node node,int max){
// System.out.println("Visiting Node " + node.value + " at depth " + max);// 判断当前节点是否是叶子节点（即左右子节点都为null）if (node.left == null && node.right == null) {cnt = Math.max(cnt, max);return; }// 遍历左子树if (node.left != null) {dfs(a[node.left.value], max + 1);}// 遍历右子树if (node.right != null) {dfs(a[node.right.value], max + 1);}}// 快速幂模版 by Dduostatic long pow(long a, long b) {if (b == 0) return 1;if (b == 1) return a;try {long result = 1;while (b > 0) {if ((b & 1) == 1) {if (result > Long.MAX_VALUE / a) return Long.MAX_VALUE;result *= a;}b >>= 1;if (b > 0) {if (a > Long.MAX_VALUE / a) return Long.MAX_VALUE;a *= a;}}return result;} catch (Exception e) {return Long.MAX_VALUE;}}
}// 数据结构模版 并查集 by Dduo
class UnionFind {private int[] parent;private int[] size;// 初始化并查集public UnionFind(int n) {parent = new int[n + 1]; // 因为编号从 1 到 N，所以数组大小是 N+1size = new int[n + 1];for (int i = 1; i <= n; i++) {parent[i] = i; // 每个元素的父节点初始为自己size[i] = 1; // 每个元素的初始大小为 1}}// 查找元素 x 所在的集合，带路径压缩优化public int find(int x) {if (parent[x] != x) {parent[x] = find(parent[x]); // 路径压缩}return parent[x];}// 合并两个集合 带按秩合并优化public void union(int x, int y) {int rootX = find(x);int rootY = find(y);if (rootX != rootY) {// 按秩合并 较小的树合并到较大的树上if (size[rootX] < size[rootY]) {parent[rootX] = rootY;size[rootY] += size[rootX];} else {parent[rootY] = rootX;size[rootX] += size[rootY];}}}// 判断 x 和 y 是否在同一个集合中public boolean connected(int x, int y) {return find(x) == find(y);}
}// 数据流快读模板(类似于C++的IOS) by Dduo
class Read{BufferedReader bf;StringTokenizer st;BufferedWriter bw;public Read(){bf=new BufferedReader(new InputStreamReader(System.in));st=new StringTokenizer("");bw=new BufferedWriter(new OutputStreamWriter(System.out));}
// 以下为输入部分:public String nextLine() throws IOException{return bf.readLine();}public String next() throws IOException{while(!st.hasMoreTokens()){st=new StringTokenizer(bf.readLine());}return st.nextToken();}public char nextChar() throws IOException{//确定下一个token只有一个字符的时候再用return next().charAt(0);}public int nextInt() throws IOException{return Integer.parseInt(next());}public long nextLong() throws IOException{return Long.parseLong(next());}public double nextDouble() throws IOException{return Double.parseDouble(next());}public float nextFloat() throws IOException{return Float.parseFloat(next());}public byte nextByte() throws IOException{return Byte.parseByte(next());}public short nextShort() throws IOException{return Short.parseShort(next());}public BigInteger nextBigInteger() throws IOException{return new BigInteger(next());}public BigDecimal nextDecimal() throws IOException{return new BigDecimal(next());}
// 以下为输出部分:public void println(int a) throws IOException{print(a);println();return;}public void print(int a) throws IOException{bw.write(String.valueOf(a));return;}public void println(String a) throws IOException{print(a);println();return;}public void print(String a) throws IOException{bw.write(a);return;}public void println(long a) throws IOException{print(a);println();return;}public void print(long a) throws IOException{bw.write(String.valueOf(a));return;}public void println(double a) throws IOException{print(a);println();return;}public void print(double a) throws IOException{bw.write(String.valueOf(a));return;}public void print(BigInteger a) throws IOException{bw.write(a.toString());return;}public void print(char a) throws IOException{bw.write(String.valueOf(a));return;}public void println(char a) throws IOException{print(a);println();return;}public void println() throws IOException{bw.newLine();return;}
// 其他调试命令：public void flush() throws IOException{
// 交互题分组调试，或者提前退出的情况下可以先运行此语句再推出bw.flush();return;}public boolean hasNext() throws IOException{
// 本地普通IDE难以使用这个方法调试，需要按照数据组flush，刷新语句:
// sc.flush()
// 调试完可删去return bf.ready();}
}