树的前序遍历、中序遍历、后序遍历,java实现
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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