106. Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

本题和之前的105题比较类似，解法可以用相同的想法来解决，代码如下：

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public TreeNode buildTree(int[] inorder, int[] postorder) {
12         return helper(postorder.length-1,0,inorder.length-1,inorder,postorder);
13     }
14     public TreeNode helper(int endPost,int startIn,int endIn,int[] inorder,int[] postorder){
15         if(endPost<0||startIn>endIn) return null;
16         TreeNode root = new TreeNode(postorder[endPost]);
17         int indexIn = 0;
18         for(int i=startIn;i<=endIn;i++){
19             if(inorder[i]==root.val){
20                 indexIn = i;
21             }
22         }
23         root.left  =helper(endPost-1-(endIn-indexIn),startIn,indexIn-1,inorder,postorder);
24         root.right = helper(endPost-1,indexIn+1,endIn,inorder,postorder);
25         return root;
26     }
27 }