Java从中序与后序遍历序列构造二叉树

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 。

示例 1:

输入：inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
输出：[3,9,20,null,null,15,7]
示例 2:

输入：inorder = [-1], postorder = [-1]
输出：[-1]

提示:

1 <= inorder.length <= 3000
postorder.length == inorder.length
-3000 <= inorder[i], postorder[i] <= 3000
inorder 和 postorder 都由 不同 的值组成
postorder 中每一个值都在 inorder 中
inorder 保证是树的中序遍历
postorder 保证是树的后序遍历

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        return buildTree2(inorder,0, inorder.length-1, postorder, 0 , postorder.length-1);
    }
    public TreeNode buildTree2(int[] inorder, int inL, int inR, int[] postorder, int postL, int postR) {
        if(postL > postR || inL > inR){
            return null;
        }
        int mid = postorder[postR];
        TreeNode root = new TreeNode(mid);
        int midIndex = inL;
        while(inorder[midIndex] != mid){
            midIndex++;
        }
        root.left = buildTree2(inorder, inL, midIndex-1, postorder, postL, postL-inL+midIndex-1);
        root.right = buildTree2(inorder,midIndex+1, inR, postorder, postL-inL+midIndex, postR-1);
        return root;
    }
}