Construct Binary Tree from Inorder and Postorder Traversal

 1 public class Solution {
 2     public TreeNode buildTree(int[] inorder, int[] postorder) {
 3         TreeNode root;
 4         if(0 == inorder.length)
 5             return null;
 6         else 
 7             return createTree(inorder, postorder, 0, inorder.length - 1, 0, postorder.length - 1);
 8         
 9     }
10     private TreeNode createTree(int []inorder, int[] postorder, int inStart, int inEnd, int postStart, int postEnd){
11         if(inStart > inEnd)
12             return null;
13         int value_root = postorder[postEnd];                //根节点的值
14         int index = 0;                                        //根节点在中序遍历中的索引值
15         for(int i = inStart; i <= inEnd; i++){                //找出根节点在终须遍历中的索引值
16             if(postorder[postEnd] == inorder[i]){
17                 index = i;
18                 break;
19             }
20         }
21         int length_leftTree = index - inStart;                //左子树的长度
22         TreeNode leftTreeRoot = createTree(inorder, postorder, inStart, index - 1, postStart, postStart + length_leftTree - 1); 
23         TreeNode rightTreeRoot = createTree(inorder, postorder, index + 1, inEnd, postStart + length_leftTree, postEnd - 1);
24         
25         TreeNode root = new TreeNode(value_root);
26         
27         root.left = leftTreeRoot;
28         root.right = rightTreeRoot;
29         
30         return root;
31     }
32 }