106. 从中序与后序遍历序列构造二叉树

106. 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]

---

class Solution {
public:
    unordered_map<int, int> map;
    int n = 0;
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        n = inorder.size();
        for (int i = 0; i < n; i++) {
            map[inorder[i]] = i;
        }
        return Pan(inorder,postorder,n-1,0,n-1);
    }
    TreeNode* Pan(vector<int>& inorder, vector<int>& postorder,int post,int start,int end) {
        if (start > end||post<0)
            return NULL;
        int flag = map[postorder[post]];
        TreeNode *root = new TreeNode(inorder[flag]);
        root->right= Pan(inorder, postorder, post-1, flag+1, end);
        root->left= Pan(inorder, postorder, post - 1+flag-end, start, flag-1);
        return root;
    }
};