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

原題：

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

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

思路：通過先序我們可以找到根節點，再到中序序列中去找到根節點，根節點的左邊便爲左子樹，右邊爲右子樹。依次遞歸下去得到答案。此次嘗試使用了迭代器來直接運算。需注意迭代器的取值用*it，vector.end()不指向任何元素。

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> &preorder, vector<int> &inorder) {
        if (preorder.empty()||inorder.empty()) return NULL;
        return make(preorder.begin(),preorder.end(), inorder.begin(), inorder.end());
    }
    
    TreeNode *make(vector<int>::iterator preleft, vector<int>::iterator preright, vector<int>::iterator inleft,vector<int>::iterator inright){
        if (preright==preleft) return NULL;
        if (inright==inleft) return NULL;
        TreeNode *root = new TreeNode(*preleft);
        vector<int>::iterator iRoot = find(inleft , inright , root->val);
        int leftsize = iRoot-inleft;
        root->right = make(preleft+leftsize+1, preright, iRoot+1, inright);
        root->left = make(preleft+1, preleft+leftsize+1, inleft, iRoot);
        return root;
    }
};

AC