【力扣刷题】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]

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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* traversal(vector<int>&inorder,int inst,int ined,vector<int>&postorder,int post,int poed){
        if(poed==post)return nullptr;
        int val = postorder[poed - 1];
        TreeNode* root = new TreeNode(val);
        if (poed-post == 1) return root;
        int cut;
        for (cut = inst; cut < ined; cut++) {
            if (inorder[cut] == val) break;
        }
        int linst = inst;
        int lined = cut;
        int rinst = cut + 1;
        int rined = ined;

        int lpost = post;
        int lpoed = post + cut - inst;
        int rpost = post + (cut - inst);
        int rpoed = poed - 1;

        root->left = traversal(inorder, linst, lined, postorder, lpost, lpoed);
        root->right = traversal(inorder, rinst, rined, postorder, rpost, rpoed);

        return root;
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        if(inorder.size()==0&&postorder.size()==0)return nullptr;
        return traversal(inorder,0,inorder.size(),postorder,0,postorder.size());
    }
};