根据前序和中序重建二叉树、后序和中序重建二叉树

题目：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。

解题思路：

像这种题首选递归来做，一个大的数组拆分成不同的区域来做。

/**
 * 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) {
        TreeNode root = reConstruct(pre,0,pre.length-1,in,0,in.length-1);
        return root;
    }
    private TreeNode reConstruct(int [] pre, int startpre,int endpre,int [] in, int startin,int endin)
   {
       if(startpre>endpre || startin>endin)
       {
           return null;
       }
       
       TreeNode root = new TreeNode (pre[startpre]);
       
       for(int i=startin; i<=endin;i++)
       {
           if(in[i]==pre[startpre])
           {
               root.left = reConstruct(pre,startpre+1,startpre+i-startin,in,startin,i-1);
               root.right = reConstruct(pre,startpre+i-startin+1,endpre,in,i+1,endin);
               
           }
       }
       return root;
       
   }
       
        
        
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        TreeNode root = recon(inorder,0,inorder.length-1,postorder,0,postorder.length-1);
        return root;
    }
    public TreeNode recon(int []inorder,int instart,int inend,int []postorder,int poststart,int postend)
    {
        if(instart>inend||poststart>postend)
            return null;
        
        TreeNode root = new TreeNode(postorder[postend]);
        
        for(int i=instart;i<=inend;i++)
        {
            if(inorder[i]==postorder[postend])
            {
                root.left = recon(inorder,instart,i-1,postorder,poststart,poststart+i-instart-1);
                root.right = recon(inorder,i+1,inend,postorder,poststart+i-instart,postend-1);
            }
        }
        return root;
    }
}