给定二叉树的先序遍历中序遍历,求后序遍历

这边博客里的代码是张晨同学写的,他自己懒得记录,我帮他记录一下

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Scanner;

/**
* Created by zhangchen(chansonzhang@163.com) on 2016/9/7.
*/
public class BinaryTree {
public static void main(String[] args){
Scanner scanner=new Scanner(System.in);
int n;
while (scanner.hasNextLine()){
n=scanner.nextInt();
List<Node<Integer>> preOrder=new ArrayList<>();
List<Node<Integer>> inOrder=new ArrayList<>();

for(int i=0;i<n;i++){
preOrder.add(new Node<Integer>(scanner.nextInt()));
}

for(int j=0;j<n;j++){
inOrder.add(new Node<Integer>(scanner.nextInt()));
}

List<Node<Integer>> postOrder=getPostOrder(preOrder,inOrder);

for(Node<Integer> node:postOrder){
System.out.print(node.getValue()+" ");
}

}
}

public static List<Node<Integer>> getPostOrder(List<Node<Integer>> preOrder,List<Node<Integer>> inOrder){
if(preOrder.size() == 0){
return new ArrayList<>();
}
Node<Integer> root=preOrder.get(0);
int rootIndexOfInOrder=inOrder.indexOf(root);
List<Node<Integer>> leftInOrder=inOrder.subList(0,rootIndexOfInOrder);
List<Node<Integer>> rightInOrder=inOrder.subList(rootIndexOfInOrder+1,inOrder.size());

List<Node<Integer>> leftPreOrder=preOrder.subList(1,1+leftInOrder.size());
List<Node<Integer>> rightPreOrder=preOrder.subList(1+leftInOrder.size(),preOrder.size());

List<Node<Integer>> postOrder=new ArrayList<>();

postOrder.addAll(getPostOrder(leftPreOrder, leftInOrder));
postOrder.addAll(getPostOrder(rightPreOrder, rightInOrder));
postOrder.add(root);

return postOrder;
}
}

class Node<T>{
private T value;
private Node<T> left;
private Node<T> right;

public Node(T value) {
this.value = value;
this.left=null;
this.right=null;
}

public T getValue() {
return value;
}

public Node<?> getLeft() {
return left;
}

public Node<?> getRight() {
return right;
}

@Override
public boolean equals(Object obj) {
return this.value.equals(((Node<T>)obj).getValue());
}
}

/*
6
1 2 4 3 6 5
4 2 1 6 3 5
4 2 6 5 3 1
*/