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 hereint 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;}};
，醒来第一眼看见的是他，然后倒头继续睡。这就是我想要的幸福。
