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。