Leetcode_106_Construct Binary Tree from Inorder and Postorder Traversal

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

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

根据中序和后序遍历确定一个二叉树，一开始打算每一个都用vector储存，结果内存超限了。
内存超限代码：

/** * Definition for a binary tree node. * 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) {        TreeNode *head = getBinaryTree(inorder, postorder);        return head;    }    TreeNode *getBinaryTree(vector<int>& inorder, vector<int>& postorder)    {        if(inorder.size() == 0) return NULL;        vector<int>::iterator it = postorder.end() - 1;        TreeNode *head = new TreeNode(*it);        if(postorder.size() == 1)        {            return head;        }        else        {            vector<int>::iterator center = find(inorder.begin(), inorder.end(),head->val);            vector<int> left_inorder;            for(vector<int>::iterator temp = inorder.begin();temp < center;temp++)            {                left_inorder.push_back(*temp);            }            vector<int> left_post;            int i = 0;            for(;i<left_inorder.size();i++)            {                left_post.push_back(postorder[i]);            }            head->left = getBinaryTree(left_inorder, left_post);            vector<int> right_inorder;            vector<int> right_post;            for(vector<int>::iterator temp = center+1;temp<inorder.end();temp++)            {                right_inorder.push_back(*temp);            }            for(int j = 0;j<right_inorder.size();j++,i++)            {                right_post.push_back(postorder[i]);            }            head->right = getBinaryTree(right_inorder, right_post);            return head;        }    }};

后来上网上查看了一下题解发现不用那么麻烦，直接在原向量上操作即可。看了思路之后还是错了好几次，主要原因就是边界的确定。
AC代码

/** * Definition for a binary tree node. * 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) {        return getBinaryTree(inorder, 0, inorder.size() -1, postorder, 0, inorder.size()-1);    }    TreeNode *getBinaryTree(vector<int>& inorder,int begin1, int end1, vector<int>& postorder, int begin2, int end2)    {         if(begin1 > end1)            return NULL;        else if(begin1 == end1)            return new TreeNode(inorder[begin1]);        TreeNode *head = new TreeNode(postorder[end2]);        int i = begin1;        for(i;i<=end1;i++)        {            if(inorder[i] == postorder[end2]) break;        }        int leftlen = i - begin1;        head->left = getBinaryTree(inorder, begin1, begin1+leftlen - 1, postorder, begin2, begin2 + leftlen -1);//注意边界        head->right = getBinaryTree(inorder, begin1+leftlen+1, end1, postorder, begin2+leftlen, end2 - 1);//注意边界        return head;    }};

解体的关键就是求出中序遍历中根结点的位置和左子树的长度。下次再做一个前序和中序遍历的题再练练这种思路。