树的前序遍历、中序遍历、后序遍历,java实现

java

1、三种遍历属于深度优先搜索(DFS,所谓前中后其实是指遍历时每个节点被访问的相对顺序。

前序遍历。节点→左孩子→右孩子  preorder

中序遍历。左孩子→节点→右孩子   inorder

后序遍历。左孩子→右孩子→节点  postorder

2、宽度优先搜索(BFS)就是从上到下,从左到右一层一层一个一个的访问

上图from leetcode

这样记忆就会很方便。

 

下面是代码:

一、前序遍历,LinkedList既可以当成栈来使用,又可以当成队列来使用

从根节点开始,每次迭代弹出当前栈顶元素,并将其孩子节点压入栈中,先压右孩子再压左孩子。输出【1,2,3,4,5】

package helloworld;

import java.util.LinkedList;

class TreeNode{

int val;

TreeNode left;

TreeNode right;

TreeNode(int x){

val=x;

}

}

public class Helloworld{

public LinkedList<Integer> preOrder(TreeNode root){

LinkedList<TreeNode> stack = new LinkedList<>();

LinkedList<Integer> output = new LinkedList<>();

if (root == null) {

return output;

}

stack.add(root);

while (!stack.isEmpty()) {

TreeNode node = stack.pollLast();

output.add(node.val);

if (node.right != null) {

stack.add(node.right);

}

if (node.left != null) {

stack.add(node.left);

}

}

return output;

}

public static void main(String[] args) {

Helloworld h=new Helloworld();

TreeNode root = new TreeNode(1);

root.left=new TreeNode(2);

root.right=new TreeNode(5);

root.left.left=new TreeNode(3);

root.left.right=new TreeNode(4);

System.out.println(h.preOrder(root));

}

}

二、中序遍历,输出[3, 2, 4, 1, 5]

方法1: 递归

package helloworld;

import java.util.ArrayList;

import java.util.List;

class TreeNode{

int val;

TreeNode left;

TreeNode right;

TreeNode(int x){

val=x;

}

}

public class Helloworld{

public List <Integer> inOrder(TreeNode root){

List < Integer > res = new ArrayList < > ();

helper(root, res);

return res;

}

public void helper(TreeNode root, List < Integer > res) {

if (root != null) {

if (root.left != null) {

helper(root.left, res);

}

res.add(root.val);

if (root.right != null) {

helper(root.right, res);

}

}

}

public static void main(String[] args) {

Helloworld h=new Helloworld();

TreeNode root = new TreeNode(1);

root.left=new TreeNode(2);

root.right=new TreeNode(5);

root.left.left=new TreeNode(3);

root.left.right=new TreeNode(4);

System.out.println(h.inOrder(root));

}

}

 方法2,基于栈遍历

package helloworld;

import java.util.*;

class TreeNode{

int val;

TreeNode left;

TreeNode right;

TreeNode(int x){

val=x;

}

}

public class Helloworld{

public List <Integer> inOrder(TreeNode root){

List < Integer > res = new ArrayList < > ();

Stack < TreeNode > stack = new Stack < > ();

TreeNode curr = root;

while (curr != null || !stack.isEmpty()) {

while (curr != null) {

stack.push(curr);

curr = curr.left;

}

curr = stack.pop();

res.add(curr.val);

curr = curr.right;

}

return res;

}

public static void main(String[] args) {

Helloworld h=new Helloworld();

TreeNode root = new TreeNode(1);

root.left=new TreeNode(2);

root.right=new TreeNode(5);

root.left.left=new TreeNode(3);

root.left.right=new TreeNode(4);

System.out.println(h.inOrder(root));

}

}

三、后序遍历,输出[3, 4, 2, 5, 1]

package helloworld;

import java.util.*;

class TreeNode{

int val;

TreeNode left;

TreeNode right;

TreeNode(int x){

val=x;

}

}

public class Helloworld{

public List <Integer> inOrder(TreeNode root){

LinkedList<TreeNode> stack = new LinkedList<>();

LinkedList<Integer> output = new LinkedList<>();

if (root == null) {

return output;

}

stack.add(root);

while (!stack.isEmpty()) {

TreeNode node = stack.pollLast();

output.addFirst(node.val);

if (node.left != null) {

stack.add(node.left);

}

if (node.right != null) {

stack.add(node.right);

}

}

return output;

}

public static void main(String[] args) {

Helloworld h=new Helloworld();

TreeNode root = new TreeNode(1);

root.left=new TreeNode(2);

root.right=new TreeNode(5);

root.left.left=new TreeNode(3);

root.left.right=new TreeNode(4);

System.out.println(h.inOrder(root));

}

}

 

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