root.left=preorder(nums, min, root.val); root.right=preorder(nums, root.val, max);returnroot; } construct bst from preorder https://www.geeksforgeeks.org/construct-bst-from-given-preorder-traversa/https://algorithms.tutorialhorizon.com/construct-binary-search-tree-from-a-given-preorder-tra...
之后在[inorder+index+1, end] [0, inorder+index-1] 分别构造右树和左树 intsearchNode(intinorder[],intinorderSize,intkey){inti;for(i=0;i<inorderSize;i++){if(key==inorder[i]){returni;}}return-1;}structTreeNode*buildTree(int*preorder,intpreorderSize,int*inorder,intinorderSize){...
self.build(node.right,right)defbstFromPreorder(self, preorder):""":type preorder: List[int] :rtype: TreeNode"""root=TreeNode(preorder.pop(0)) self.build(root,preorder)returnroot
classSolution(object):defbstFromPreorder(self,preorder):""" :type preorder: List[int] :rtype: TreeNode """root=TreeNode(preorder[0])stack=[root]foriinpreorder[1:]:i=TreeNode(i)ifi.val<stack[-1].val:stack[-1].left=i stack.append(i)else:whilestackandstack[-1].val<i.val:last...
// Helper function to perform the preorder traversal on a BST void preorder(Node* root) { if (root == nullptr) { return; } preorder(root->left); cout << root->data << " "; preorder(root->right); } // Function to insert a BST node at the front of a doubly linked list vo...
fun bstFromPreorder(preorder: IntArray): TreeNode? { if (preorder == null || preorder.isEmpty()) { return null } val size = preorder.size val root = TreeNode(preorder[0]) for (i in 1 until size) { help(root, preorder[i]) } return ...
node->right =bstFromPreorder(right);returnnode; } }; Github 同步地址: https://github.com/grandyang/leetcode/issues/1008 类似题目: Construct Binary Tree from Preorder and Postorder Traversal Construct Binary Tree from Inorder and Postorder Traversal ...
1/**2* Definition for a binary tree node.3* public class TreeNode {4* int val;5* TreeNode left;6* TreeNode right;7* TreeNode(int x) { val = x; }8* }9*/10classSolution {11publicTreeNode bstFromPreorder(int[] preorder) {12if(preorder ==null|| preorder.length == 0){13...
Given an array of integers preorder, which represents the preorder traversal of a BST (i.e., binary search tree), construct the tree and returnits root. It is guaranteed that there is always possible to find a binary search tree with the given requirements for the given test cases. ...
TreeNode* bstFromPreorder(vector<int>&preorder) {returndfs(0, preorder.size() -1, preorder); } }; 2. O(N) classSolution {public:inti =0; TreeNode* bstFromPreorder(vector<int>& A,intbound =INT_MAX) {if(i == A.size() || A[i] > bound)returnnullptr; ...