LeetCode Top 100 Liked Questions 105. Construct Binary Tree from Preorder and Inorder Traversal (Java版; Medium) 题目描述 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, given preorder ...
https://leetcode.com/problems/binary-tree-postorder-traversal 问题:二叉树后序遍历。 思路: publicList<Integer>postorderTraversal(TreeNoderoot){LinkedList<Integer>results=newLinkedList<>();if(root==null)returnresults;Stack<TreeNode>stack=newStack<>();stack.push(root);while(!stack.isEmpty()){Tree...
1classSolution {2public:3vector<int> preorderTraversal(TreeNode *root) {4vector<int>result;5TreeNode *p =root;6while(p) {7if(!p->left) {8result.push_back(p->val);9p = p->right;10}else{11TreeNode *q = p->left;12while(q->right && q->right !=p) {13q = q->right;14}1...
LeetCode Top 100 Liked Questions 236. Lowest Common Ancestor of a Binary Tree (Java版; Medium) 题目描述 Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree. According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined betw...
Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest l
图片来源:https://leetcode.com/articles/binary-tree-right-side-view/ 对于时间复杂度来说,基本上都是O(n),因为要访问所有的点。 对于空间复杂度来说,BFS取决于扫描过程中每层的node数,就是树的宽度,而DFS取决于扫描过程中树的深度。最坏情况两个都是O(n)。
return dfs(root.Left, low, &root.Val) && dfs(root.Right, &root.Val, high) } 题目链接: Validate Binary Search Tree: leetcode.com/problems/v 带因子的二叉树: leetcode.cn/problems/va LeetCode 日更第 209 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满 ...
英文网站:199. Binary Tree Right Side View 中文网站:199. 二叉树的右视图 问题描述 Given therootof a binary tree, imagine yourself standing on theright sideof it, returnthe values of the nodes you can see ordered from top to bottom.
已解答 0% 通过率 击败用户 0% 击败用户 0% 击败用户 0% 0尝试中 0次提交 0尝试中 0尝试中 0尝试中 简单 0/52 中等 0/149 困难 0/23 中等 96. 不同的二叉搜索树 中等 98. 验证二叉搜索树 中等 99. 恢复二叉搜索树 中等 100. 相同的树 ...
解释: 节点 2 和节点 4 的最近公共祖先是 2, 因为根据定义最近公共祖先节点可以为节点本身。 说明: 所有节点的值都是唯一的。 p、q 为不同节点且均存在于给定的二叉搜索树中。 来源:力扣(LeetCode) 链接:https://leetcode.cn/problems/... 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明...