Lintcode前序遍历和中序遍历树构造二叉树

样例

给出中序遍历：[1,2,3]和前序遍历：[2,1,3]. 返回如下的树:

  2 / \1   

/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     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) {        // write your code here                int len1 = preorder.size();        int len2 = inorder.size();        if (len1 <= 0 || len2 <= 0) {            return nullptr;        }        return  dp(preorder,inorder);                }    TreeNode *dp(vector<int> &porder, vector<int> &iorder) {         int len1 = porder.size();        int len2 = iorder.size();        if (len1 <= 0 || len2 <= 0) {            return nullptr;        }            TreeNode *root = new TreeNode();        root->val=porder[0];        if (len1 ==1 || len2 ==1) {            return root;        }        int r=0;        for (int i = 0; i < len1 ; ++i) {            if (iorder[i] == root->val) {                r = i;                break;            }        }        vector<int> linorder;        vector<int> rinorder;        vector<int> rporder;        vector<int> lporder;         for (int i = 0; i < r ; ++i) {             linorder.push_back(iorder[i]);         }         for (int i = r+1; i < len2 ; ++i) {             rinorder.push_back(iorder[i]);         }          for (int i = 1; i <= r ; ++i) {             lporder.push_back(porder[i]);         }          for (int i = r+1; i <= len1; ++i) {             rporder.push_back(porder[i]);         }                 root->left =dp(lporder,linorder);         root->right =dp(rporder,rinorder);         return root;    }};