1. 给定两个整数数组 inorder 和 postorder ，其中 inorder 是二叉树的中序遍历， postorder 是同一棵树的后序遍历，请你构造并返回这颗 二叉树 。
    3. 示例 1:
    4. 输入：inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
    5. 输出：[3,9,20,null,null,15,7]

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    1. /**
    2.  * Definition for a binary tree node.
    3.  * public class TreeNode {
    4.  *     int val;
    5.  *     TreeNode left;
    6.  *     TreeNode right;
    7.  *     TreeNode() {}
    8.  *     TreeNode(int val) { this.val = val; }
    9.  *     TreeNode(int val, TreeNode left, TreeNode right) {
    10.  *         this.val = val;
    11.  *         this.left = left;
    12.  *         this.right = right;
    13.  *     }
    14.  * }
    15.  */
    16. class Solution {
    17.     Map<Integer,Integer> myMap = new HashMap();
    18.     public TreeNode buildTree(int[] inorder, int[] postorder) {
    19.         for(int i=0; i < inorder.length; i++)  myMap.put(inorder[i],i);
    20.         return backTracking(inorder,postorder, 0, postorder.length, 0, inorder.length);
    21.     }
    22.     public TreeNode backTracking(int[] inorder, int[] postorder,
    23.             int pStart, int pEnd, int iStart, int iEnd){
    25.         if(pStart == pEnd) return null;
    27.         int rootVal = postorder[pEnd-1];
    28.         TreeNode root = new TreeNode(rootVal);
    29.         int inorderIdx = myMap.get(rootVal);
    31.         int leftNum = inorderIdx - iStart;
    33.         root.left = backTracking(inorder, postorder,
    34.                                  pStart, pStart+leftNum, iStart, inorderIdx);
    35.         root.right = backTracking(inorder, postorder,
    36.                                   pStart+leftNum, pEnd-1, inorderIdx+1, iEnd);
    37.         return root;
    38.     }
    39. }