106 Construct Binary Tree from Inorder and Postorder Traversal 从中序与后序遍历序列构造二叉树

给定一棵树的中序遍历与后序遍历，依据此构造二叉树。
注意:
你可以假设树中没有重复的元素。
例如，给出
中序遍历 = [9,3,15,20,7]
后序遍历 = [9,15,7,20,3]
返回如下的二叉树：
    3
   /
  9  20
    / 
   15   7
详见：https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/

Java实现：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder==null||inorder.length==0||postorder==null||postorder.length==0||inorder.length!=postorder.length){
            return null;
        }
        return buildTree(inorder,0,inorder.length-1,postorder,0,postorder.length-1);
    }
    private TreeNode buildTree(int[] inorder,int startIn,int endIn,int[] postorder,int startPost,int endPost){
        if(startIn>endIn||startPost>endPost){
            return null;
        }
        TreeNode root=new TreeNode(postorder[endPost]);
        for(int i=startIn;i<=endIn;++i){
            if(inorder[i]==postorder[endPost]){
                root.left=buildTree(inorder,startIn,i-1,postorder,startPost,startPost+i-startIn-1);
                root.right=buildTree(inorder,i+1,endIn,postorder,startPost+i-startIn,endPost-1);
            }
        }
        return root;
    }
}