剑指offer：重建二叉树

题目描述

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。

方法1：利用Arrays.copyOfRange方法来实现

【运行时间：260ms   占用内存：22048k】

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */import java.util.*;public class Solution {    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {        if(pre.length == 0||in.length == 0){return null;}TreeNode node = new TreeNode(pre[0]);for(int i = 0; i < in.length; i++){if(pre[0] == in[i]){node.left = reConstructBinaryTree(Arrays.copyOfRange(pre, 1, i+1), Arrays.copyOfRange(in, 0, i));node.right = reConstructBinaryTree(Arrays.copyOfRange(pre, i+1, pre.length), Arrays.copyOfRange(in, i+1,in.length));}}return node;    }}

方法2：用二叉树的递归来实现

【运行时间：336ms  占用内存：19440k】

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {       return reConBTree(pre, 0, pre.length - 1, in, 0, in.length - 1);    }        public TreeNode reConBTree(int[] pre, int preleft, int preright, int[] in, int inleft, int inright) {// 当到达边界条件时候返回nullif (preleft > preright || inleft > inright)return null;// 新建一个TreeNodeTreeNode root = new TreeNode(pre[preleft]);// 对中序数组进行输入边界的遍历for (int i = inleft; i <= inright; i++) {if (pre[preleft] == in[i]) {// 重构左子树，注意边界条件root.left = reConBTree(pre, preleft + 1, preleft + i - inleft, in, inleft, i - 1);// 重构右子树，注意边界条件root.right = reConBTree(pre, preleft + i + 1 - inleft, preright, in, i + 1, inright);}}return root;}}