Construct Binary Tree from Inorder and Postorder Traversal

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

根据树的中序（左根右）和后序遍历（左右根）构造一棵二叉树

参考：http://www.cnblogs.com/remlostime/archive/2012/10/29/2744738.html

后序遍历最后一个元素是根节点，通过根节点将中序遍历数组分为两个数组，分别是左子树和右子树节点的集合，在进行递归调用

这里用的下标，节约空间

public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        TreeNode root;
        if(0 == inorder.length)
            return null;
        else 
            return createTree(inorder, postorder, 0, inorder.length - 1, 0, postorder.length - 1);
        
    }
    private TreeNode createTree(int []inorder, int[] postorder, int inStart, int inEnd, int postStart, int postEnd){
        if(inStart > inEnd)
            return null;
        int value_root = postorder[postEnd];                //根节点的值
        int index = 0;                                        //根节点在中序遍历中的索引值
        for(int i = inStart; i <= inEnd; i++){                //找出根节点在终须遍历中的索引值
            if(postorder[postEnd] == inorder[i]){
                index = i;
                break;
            }
        }
        int length_leftTree = index - inStart;                //左子树的长度
        TreeNode leftTreeRoot = createTree(inorder, postorder, inStart, index - 1, postStart, postStart + length_leftTree - 1); 
        TreeNode rightTreeRoot = createTree(inorder, postorder, index + 1, inEnd, postStart + length_leftTree, postEnd - 1);
        
        TreeNode root = new TreeNode(value_root);
        
        root.left = leftTreeRoot;
        root.right = rightTreeRoot;
        
        return root;
    }
}