Construct Binary Tree from Inorder and Postorder Traversal

通过中序和后序建立一个二叉树（或前序和中序），递归解法三点要注意的：

1.不要new新的vector来存储左子树和右子树的中序和后序序列，这样做容易Memory Limit Exceeded

2.构造新函数的时候参数值要记得用引用，要不然也会Memory Limit Exceeded

3.弄清楚左子树（右子树）的中序（后序）的起始点和终止点的下标

/** * 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.size() == 0 || postorder.size() == 0 || inorder.size() != postorder.size()){            return NULL;        }                return buildTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);    }        TreeNode *buildTree(vector<int> &inorder, int in_start, int in_end, vector<int> &postorder, int post_start, int post_end){        if(in_start > in_end || post_start > post_end)            return NULL;                    int i, j;        int val = postorder.at(post_end);        TreeNode *root = new TreeNode(val);        for(i = in_start; i <= in_end; i++){            if(inorder.at(i) == val)                break;        }                TreeNode *left = buildTree(inorder, in_start, i - 1, postorder, post_start, post_start + i - in_start - 1);        TreeNode *right = buildTree(inorder, i + 1, in_end, postorder, post_start + i - in_start, post_end - 1);        root -> left = left;        root -> right = right;        return root;    }};