classSolution {int[] preorder;intidx = 0;intn;publicTreeNode bstFromPreorder(int[] preorder) {this.preorder =preorder; n=preorder.length;returnhelper(Integer.MIN_VALUE, Integer.MAX_VALUE); }publicTreeNode helper(intlower,intupper){if(idx == n)returnnull;intval =preorder[idx];if(val ...
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; TreeNode* root =new...
dfs(preorder,root->right,i+1,right); break; } } if(i==right+1){ root->left=newTreeNode(preorder[left]); dfs(preorder,root->left,left+1,i-1); } } } TreeNode*bstFromPreorder(vector<int>&preorder) { intlen=preorder.size(); TreeNode*root=newTreeNode(preorder[0]); dfs(pre...
classSolution {publicTreeNode bstFromPreorder(int[] preorder) {if(preorder ==null|| preorder.length ==0) {returnnull; } Stack<TreeNode> stack =newStack<>(); TreeNode root=newTreeNode(preorder[0]); stack.push(root);for(inti =1; i < preorder.length; i++) { TreeNode node=newT...
Let's do the dry run and see how it creates the BST from the level order traversal. First, create the root from the very first element of the level order traversal Now insert the next element 2 and that will go towards left of 3 ...
- 测试代码:funcConstructBinaryTreeFromInorderandPostorderTraversal(){letinorder=[4,2,5,1,6,3,7]letpostorder=[4,5,2,6,7,3,1]letroot=Solution().buildTree(inorder,postorder)print(root??"二叉树为空",CreateBinaryTree().isValidBST(root:root))//额外判断一下一颗二叉树是否为二叉查找树, 并...
Inorder traversal is the sorted array so now we can say the problem is reduced to construct a BST from the sorted array. Let me know your thoughts. If agree with my thought process then u can use the following link http://www.geeksforgeeks.org/sorted-array-to-balanced-bst/ commented...
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-traversal-using-recursion/Method1 ( O(n^2) ti...
node->left =bstFromPreorder(left); 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 ...
leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal 中前序构造BST,Givenpreorderandinordertraversalofatree,constructthebinarytree.本题就是根据前序遍历和中序遍历的结果还原一个二叉树。题