lc-106

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

//leetcode submit region begin(Prohibit modification and deletion)
/**
 * 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 createSubTree(inorder, postorder, 0, inorder.length - 1, 0, postorder.length - 1);
    }


    private TreeNode createSubTree(int[] inorder, int[] postorder, int inStart, int inEnd, int postStart, int postEnd) {
        if (inStart > inEnd) {
            return null;
        }
        if (inEnd == inStart && postEnd == postStart) {
            return new TreeNode(inorder[inStart]);
        }

        int root = postorder[postEnd];
        int inIndex = inStart;
            for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == root) {
                inIndex = i;
                break;
            }
        }

        int len = inIndex - inStart;
        return new TreeNode(root, createSubTree(inorder, postorder, inStart, inIndex - 1, postStart, postStart + len - 1),
        createSubTree(inorder, postorder, inIndex + 1, inEnd, postStart + len, postEnd - 1));
    }
}
//leetcode submit region end(Prohibit modification and deletion)