Construct Binary Tree from Preorder and Inorder Traversal
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
代码：
1 TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) { 2 // IMPORTANT: Please reset any member data you declared, as 3 // the same Solution instance will be reused for each test case. 4 int len = preorder.size(); 5 if(len == 0) 6 return NULL; 7 TreeNode *root = new TreeNode(preorder[0]); 8 int i; 9 for(i = 0; i < len; i++){ 10 if(inorder[i] == preorder[0]) 11 break; 12 } 13 vector<int> lpreorder(preorder.begin()+1, preorder.begin()+i+1); 14 vector<int> linorder(inorder.begin(), inorder.begin()+i); 15 vector<int> rpreorder(preorder.begin()+i+1, preorder.end()); 16 vector<int> rinorder(inorder.begin()+i+1, inorder.end()); 17 vector<int>().swap(preorder); 18 vector<int>().swap(inorder); 19 root->left = buildTree(lpreorder, linorder); 20 root->right = buildTree(rpreorder, rinorder); 21 return root; 22 }
