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

来源：互联网发布：百度贴吧营销软件编辑：程序博客网时间：2024/06/04 18:03

原题：

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：和上一题一样，后续我们可以通过最后一个值得到根的值，同样可以通过定位根的值得到左右子树的子集，递归求解即可。

代码：

/** * 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;    }};

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