【Leetcode】Construct Binary Tree from Inorder and Postorder Traversal

题目链接：https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

题目：

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

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

思路：

easy

算法：

 public TreeNode buildTree(int[] inorder, int[] postorder) {
     if (inorder == null || inorder.length == 0 || postorder == null || postorder.length == 0) {
         return null;
     }

     // 在前序排列中，从头开始查找定位： 第一个既在后序也在中序的元素位置，以及它在后序中的位置
     int start = -1, pindex = -1;// start是后序中的位置 index是中序的位置
     for (int i = postorder.length - 1; i >= 0; i--) {
         int tmp = search(inorder, postorder[i]);
         if (tmp >= 0) {
             pindex = tmp;
             start = i;
             break;
         }
     } // ====end
     TreeNode root = new TreeNode(postorder[start]);
     if (inorder.length == 1) {
         return root;
     }
     // 根据index划分中序排列
     int leftInorder[] = Arrays.copyOfRange(inorder, 0, pindex);
     int rightInorder[] = Arrays.copyOfRange(inorder, pindex + 1, inorder.length);

     int newpreorder[] = Arrays.copyOfRange(postorder, 0, start);

     root.left = buildTree(leftInorder, newpreorder);
     root.right = buildTree(rightInorder, newpreorder);
     return root;
 }

 public int search(int nums[], int target) {
     for (int i = 0; i < nums.length; i++) {
         if (nums[i] == target)
             return i;
     }
     return -1;
 }