6* public var right: TreeNode?7* public init(_ val: Int) {8* self.val = val9* self.left = nil10* self.right = nil11* }12* }13*/14classSolution {15vari =01617func bstFromPreorder(_ preorder: [Int]) -> TreeNode?
1008. 先序遍历构造二叉树 # Definition for a binary tree node.# class TreeNode(object):# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclassSolution(object):defbstFromPreorder(self,preorder):ifpreorder:root=TreeNode(preord...
<pre language="typescript" code_block="true"> public TreeNode bstFromPreorder2(int[] preorder) { return bstFromPreorder(preorder, 0, preorder.length - 1); } private TreeNode bstFromPreorder(int[] preorder, int start, int end) { if (start > end) { return null; } TreeNode root...
查询index部分可以使用Map做查询,建议和下一道题 leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal 中后序构造BST和leetcode 449. Serialize and Deserialize BST 二叉搜索树BST的序列化和反序列化一起学习。 建议和leetcode 87. Scramble String 字符串拼凑 && DFS深度优先搜索 和 leetco...
LeetCode树专题(遍历,BST,Trie)(未完成) 层次遍历 使用BFS 进行层次遍历。不需要使用两个队列来分别存储当前层的节点和下一层的节点,因为在开始遍历一层的节点时,当前队列中的节点数就是当前层的节点数,只要控制遍历这么多节点数,就能保证这次遍历的都是当前层的节点。
☕ 题目:LeetCode144. 二叉树的前序遍历 (https://leetcode-cn.com/problems/binary-tree-preorder-traversal/) ❓ 难度:简单 📕 描述:给你二叉树的根节点root,返回它节点值的前序遍历。 题目示例 💡 思路: 递归法前序遍历 我们前面看了递归三要素,接下来我们开始用递归法来进行二叉树的前序遍历: ...
bool: self.nodeValueList = [] if root is None: return True if root.left is None and root.right is None: return True self.inOrder(root) for i in range(len(self.nodeValueList) - 1): if self.nodeValueList[i] >= self.nodeValueList[i + 1]: self.isBST = False return self.isBST ...
333 Largest BST Subtree $ 26.50% Medium 332 Reconstruct Itinerary 23.40% Medium 331 Verify Preorder Serialization of a Binary Tree 31.50% Medium 330 Patching Array 28.80% Medium 329 Longest Increasing Path in a Matrix 29.50% Medium 328 Odd Even Linked List 37.80% Easy 327 Count of Range Sum ...
思路:中序遍历时,判断当前节点是否大于中序遍历的前一个节点,如果大于,说明满足 BST,继续遍历;否则直接返回 false class Solution { long pre = Long.MIN_VALUE; public boolean isValidBST(TreeNode root) { if (root == null) return true; // 访问左子树 if (!isValidBST(root.left)) return false;...
0331 Verify Preorder Serialization of a Binary Tree Go 40.4% Medium 0332 Reconstruct Itinerary 36.7% Medium 0333 Largest BST Subtree 35.8% Medium 0334 Increasing Triplet Subsequence 40.0% Medium 0335 Self Crossing 28.0% Hard 0336 Palindrome Pairs 33.7% Hard 0337 House Robber III Go 50.6%...