剑指offer-重建二叉树

 1 public class Solution {//树 my
 2     public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
 3         return reConBTreeIndex(pre,0,pre.length-1,in,0,in.length-1);
 4     }
 5     private TreeNode reConBTreeIndex(int[] pre,int preLeft,int preRight,int[] in,int inLeft,int inRight){
 6         //和1思路完全一样，代码更简洁
 7         if(preRight<preLeft||inLeft>inRight){
 8             return null;
 9         }
10         else{
11             //System.out.println(pre[preLeft]+"--"+pre[preRight]+"   "+in[inLeft]+"--"+in[inRight]);
12             TreeNode node = new TreeNode(pre[preLeft]);
13             for (int i = inLeft; i <= inRight; i++) {
14                 if(in[i]==pre[preLeft]){
15                     node.left = reConBTreeIndex(pre,preLeft+1,i-inLeft+preLeft,in,inLeft,i-1);
16                     node.right = reConBTreeIndex(pre,preRight-inRight+i+1,preRight,in,i+1,inRight);
17                 }
18             }
19             return node;
20         }
21     }
22 }