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-traversal-using-recursion/Method1 ( O(n^2) time complexity ) The first element of preorder traversal is ...
self.build(node.right,right)defbstFromPreorder(self, preorder):""":type preorder: List[int] :rtype: TreeNode"""root=TreeNode(preorder.pop(0)) self.build(root,preorder)returnroot
Example 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.v...
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 ...
// Helper function to perform the preorder traversal on a BST void preorder(Node* root) { if (root == nullptr) { return; } cout << root->data << " "; preorder(root->left); preorder(root->right); } // Recursive function to push nodes of a given binary search tree into a ...
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...
The values ofpreorderare distinct. classSolution {publicTreeNode bstFromPreorder(int[] preorder) {if(preorder ==null|| preorder.length ==0) {returnnull; } Stack<TreeNode> stack =newStack<>(); TreeNode root=newTreeNode(preorder[0]); ...
Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example,
* preorder:root->left->right,so the first element of preorder array is the root of tree **/fun bstFromPreorder(preorder: IntArray): TreeNode?{if(preorder ==null||preorder.isEmpty()) {returnnull} val size=preorder.size val root= TreeNode(preorder[0])for(i in 1until size) { ...