已知先序遍历和中序遍历，求后序遍历

Posted 2020-08-29 hellozay

描述

先序序列{1,2,4,8,9,5,10,3,6,7};
中序序列{8,4,9,2,10,5,1,6,3,7};
求后序序列。

 

代码

class Node{
    public int data;
    public Node left;
    public Node right;
    public Node(int data){
        this.data=data;
        this.left=null;
        this.right=null;
    }
}

public class BinaryTree {
    private Node root;
    public BinaryTree(){
        root=null;
    }

    /*
    * 初始化树
    * */
    public void initTree(int[] preOrder,int[] inOrder){
        root=buildTree(preOrder,0,preOrder.length-1,inOrder,0,inOrder.length-1);
    }
    /*
    * 递归法，建立左右子树
    * */
    public Node buildTree(int[] preOrder,int start1,int end1,int[] inOrder,int start2,int end2){
        if(start1>end1 || start2>end2) return null;

        //先序序列中首字符则为根结点
        int rootData=preOrder[start1];
        Node head=new Node(rootData);
        //从中序序列inOrder的[start2,end2]中找到该结点rootData的下标，即可确定左子树和右子树的边界
        int rootIndex=findIndexInArray(inOrder,rootData,start2,end2);
        //确定左子树有多少结点，从而能确定先序序列中的左子树部分。
        int leftLen=rootIndex-start2;
        //构建左子树
        Node left=buildTree(preOrder,start1+1,start1+leftLen,inOrder,start2,rootIndex-1);
        //构建右子树
        Node right=buildTree(preOrder,start1+leftLen+1,end1,inOrder,rootIndex+1,end2);

        head.left=left;
        head.right=right;
        return head;
    }

    /*
    * 从数组a的[begin,end]中找到x的下标
    * */
    public int findIndexInArray(int[] a,int x,int begin,int end){
        for(int i=begin;i<=end;i++){
            if(x==a[i]) return i;
        }
        return -1;
    }
    
    /*
    * 后序遍历
    * */
    public void postOrder(){
        postOrder(root);
    }
    public void postOrder(Node node){
        if(node==null) return;
        postOrder(node.left);
        postOrder(node.right);
        System.out.print(node.data+" ");
    }

    public static void main(String[] args) {
        BinaryTree bitree=new BinaryTree();
        int[] preOrder={1,2,4,8,9,5,10,3,6,7};
        int[] inOrder={8,4,9,2,10,5,1,6,3,7};
        bitree.initTree(preOrder,inOrder);
        bitree.postOrder();
    }
}