leetcode日记（93）从中序与后序遍历序列构造二叉树

和上一题类似，要从后往前遍历后序遍历，并且先右后左。

/**
 * 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:
    vector<int> postorder;
    unordered_map<int,int> index;
    int n;
    TreeNode* dg(int min,int max){
        if(min>max||n<0) return nullptr;
        TreeNode* tree=new TreeNode(postorder[n]);
        if(max==min) return tree;
        int i=index[postorder[n]];
        if(max!=i){
            n--;
            tree->right=dg(i+1,max);
        }
        if(min!=i){
            n--;
            tree->left=dg(min,i-1);
        }
        return tree;
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        this->postorder=postorder;
        for(int i=0;i<inorder.size();i++){
            index[inorder[i]]=i;
        }
        n=postorder.size()-1;
        return dg(0,postorder.size()-1);
    }
};