二叉树重建-递归的终止条件不要忘记，要有出口啊。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private int[] pre;
    private int[] in;
    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
        /*
        要注意一下，
        */
        this.pre=pre;
        this.in=in;
        if(pre==null||in==null){
            return null;
        }
        if(pre.length==0||in.length==0){
            return null;
        }
       TreeNode treeNode=reBuildBinaryTree(0,pre.length-1,0,in.length-1);

       return treeNode;


    }
    private TreeNode reBuildBinaryTree(int st1,int ed1,int st2,int ed2){
        if(st1==ed1||st2==ed2){
            return new TreeNode(pre[st1]);
        }
        if(st1>ed1||st2>ed2){
            return null;
        }
        TreeNode root=new TreeNode(pre[st1]);
        int par=st2;
        for(int i=st2;i<=ed2;i++){
            if(in[i]==pre[st1]){
                par=i;
                break;
            }
        }
        int leftsize=par-st2;
        int rightsize=ed2-par;
        root.left=reBuildBinaryTree(st1+1,st1+leftsize,st2,par-1);
        root.right=reBuildBinaryTree(st1+leftsize+1,ed1,par+1,ed2);
        return root;
    }

}