唯一确认一棵二叉树

这个要通过递归来做，

如：给一个前序遍历{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}

解题思路：

  1.先通过前序序列确认根节点（即数组的第一个），再通过中序序列确认左右子树

   2.分别对左右子树来进行递归，直到左右子树变为叶子节点为止




代码如下：

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





public class Solution {

    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {

        TreeNode tree=new TreeNode(pre[0]);

        if(pre.length>1){

            int count=0;

            for(int i=0;i<in.length;i++)

                {

                  if(in[i]==pre[0])break;

                  count++;               

                }

            if(count>0){

            int pre1[]=new int[count];

            int in1[]=new int[count];

            for(int i=1;i<=count;i++)

                {

                  pre1[i-1]=pre[i];

               }

             for(int i=0;i<count;i++)

                 {

                  in1[i]=in[i];  

              }

              tree.left=reConstructBinaryTree(pre1,in1);

            }

            if(pre.length-count-1>0){

             int pre2[]=new int[pre.length-count-1];

             int in2[]=new int[pre.length-count-1];

              for(int i=count+1;i<pre.length;i++)

                  {

                     pre2[i-count-1]=pre[i];

                     in2[i-count-1]=in[i];

                 }

                tree.right=reConstructBinaryTree(pre2,in2);

            }

        }

        return tree; 

    }

}