实现二叉树先序，中序，后序

题目描述：

分别按照二叉树先序，中序，后序打印所有的节点

输入：

{1，2，3}

返回值

[[1,2,3],[2,1,3],[2,3,1]]

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 * }
 */

public class Solution {
    /**
     *
     * @param root TreeNode类 the root of binary tree
     * @return int整型二维数组
     */
    private int preIndex = 0;
    private int inIndex = 0;
    private int postIndex = 0;
    
    public int[][] threeOrders (TreeNode root) {
        // write code here
        // 前序
        int size = getNums(root);
        
        int[][] res = new int[3][size];
        preOrder(root,res);
        inOrder(root,res);
        postOrder(root,res);
        return res;
    }
    public void postOrder(TreeNode root,int[][] res){
        if(root == null){
            return;
        } 
        postOrder(root.left,res);
        postOrder(root.right,res);
        res[2][postIndex++] = root.val;
    }
    
    public void inOrder(TreeNode root,int[][] res){
        if(root == null){
            return;
        } 
        inOrder(root.left,res);
        res[1][inIndex++] = root.val;
        inOrder(root.right,res);
    }
    
    public void preOrder(TreeNode root,int[][] res){
        if(root == null){
            return;
        } 
        res[0][preIndex++] = root.val;
        preOrder(root.left,res);
        preOrder(root.right,res);
    }
    
    
    public int getNums(TreeNode root){
        if (root == null ){
            return 0;
        }
            /*
            root  TreeNode@7d4991ad
            root.left  TreeNode@28d93b30
            root.right  TreeNode@1b6d3586

            root  TreeNode@28d93b30
            root.left  null
            root.right  null

            root  TreeNode@1b6d3586
            root.left  null
            root.right  null

        */
        System.out.println("root  " + root);
        System.out.println("root.left  " + root.left);
        System.out.println("root.right  " + root.right);
        return 1 + getNums(root.left) + getNums(root.right);
    }
}