重构二叉树

输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。

思路：找到前序数组每一个元素在中序数组中的下标，该下标的值是子树中的根节点的值，其左右子树可以按照中序的下标的前后划分。如此循环迭代直至数组中的每个元素都粘到树上面。下面是我的实现（为了验证重建的正确性，我将重建的树又按照前序遍历和中序遍历数出来，可供校验）

package com.example.test;


public class TreeRebuild {

	public class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;
		TreeNode(int x) {
			val = x;
		}
	}

	public static void main(String[] args) {
		int []pre={1,2,4,7,3,5,6,8};
		int[] in = {4,7,2,1,5,3,8,6};
		TreeRebuild treeRebuild =new TreeRebuild();
		TreeNode tree=treeRebuild.reConstructBinaryTree(pre, in);
		System.out.println("树的前序遍历：");
		treeRebuild.printByPre(tree);
		System.out.println("\n树的中序遍历：");
		treeRebuild.printByOrder(tree);

	}

	public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
		TreeNode tree=reConstructBinaryTree(pre, 0, pre.length, in, 0, in.length);
		return tree;
	}
	
	public TreeNode reConstructBinaryTree(int[] pre, int pstart,int pend,int[] in, int istart,int iend){
		
		if(pend-pstart<=0||iend-istart<=0)
			return null;
		TreeNode treeNode=new TreeNode(pre[pstart]);
		for(int i=0;i<iend;i++){
			if(in[i]==pre[pstart]){
				treeNode.left=reConstructBinaryTree(pre, pstart+1,pstart+1+i-istart, in, istart, i);
				treeNode.right=reConstructBinaryTree(pre, pstart+1+i-istart, pend, in, i+1, iend);
			}
		}
		return treeNode;
	}
	
	public void printByPre(TreeNode treeNode){
		if(treeNode==null)
			return;
		System.out.print(treeNode.val+" ");
		printByPre(treeNode.left);
		printByPre(treeNode.right);
	}
	public void printByOrder(TreeNode treeNode){
		if(treeNode==null)
			return;
		printByOrder(treeNode.left);
		System.out.print(treeNode.val+" ");
		printByOrder(treeNode.right);
	}

}