另一种递归构造二叉树，不同于剑指offer

//Definition of TreeNode:
class TreeNode {
public:
    int val;
    TreeNode *left, *right;
    TreeNode() {
        this->val = NULL;
    }
    TreeNode(int val) {
        this->val = val;
        this->left = this->right = NULL;
    }
};

class Solution {
    /**
    *@param preorder : A list of integers that preorder traversal of a tree
    *@param inorder : A list of integers that inorder traversal of a tree
    *@return : Root of a tree
    */
public:
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        // recursive return
        TreeNode *root;
        // 迭代到没有再返回
        if (preorder.size() == 0 || inorder.size() == 0) {
            root = NULL;
            return NULL;
        }

        // write your code here
        int rootValue = preorder[0];
        root = new TreeNode(rootValue);

        int index;
        for (int i = 0; i < inorder.size(); i++) {
            if (inorder[i] == rootValue) {
                index = i;
                break;
            }
        }
　　　　 //黑科技构造vector，迭代器可以这么用
        vector<int> lchild_preorder(preorder.begin() + 1, preorder.begin() + 1 + index);
        vector<int> lchild_inorder(inorder.begin(), inorder.begin() + index);
        int lchild_len = lchild_preorder.size();

        vector<int> rchild_preorder(preorder.begin() + index + 1, preorder.end());
        vector<int> rchild_inorder(inorder.begin() + index + 1, inorder.end());
        int rchild_len = rchild_preorder.size();

        if (root!=NULL) {
            //root->val = rootValue;
            root->left = buildTree(lchild_preorder, lchild_inorder);
            root->right = buildTree(rchild_preorder, rchild_inorder);
        }
        return root;
    }
};

http://www.lintcode.com/zh-cn/problem/construct-binary-tree-from-preorder-and-inorder-traversal/

accepted