【Leetcode長征系列】Construct Binary Tree from Inorder and Postorder Traversal

原題：

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

思路：和上一題一樣，後續我們可以通過最後一個值得到根的值，同樣可以通過定位根的值得到左右子樹的子集，遞歸求解即可。

代碼：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        if(inorder.empty()||postorder.empty()) return NULL;
        return make(postorder.begin(), postorder.end(), inorder.begin(), inorder.end());
    }
    
    TreeNode *make(vector<int>::iterator pFirst, vector<int>::iterator pEnd, vector<int>::iterator iFirst, vector<int>::iterator iEnd){
        if(pFirst==pEnd||iFirst==iEnd) return NULL;
        TreeNode *root = new TreeNode(*(pEnd-1));
        vector<int>::iterator iRoot = find(iFirst, iEnd, root->val);
        int rightsize = iEnd-iRoot;
        root->left = make(pFirst, pEnd-rightsize, iFirst, iRoot);
        root->right = make(pEnd-rightsize-1, pEnd-1, iRoot+1, iEnd);
        return root;
    }
};

AC