Maximum Depth of Binary Tree: https://leetcode.com/problems/maximum-depth-of-binary-tree/ 这道题应该算是leetcode中最简单的一道binary tree的题了。只要想清楚递归的关系就好。 View Code Unique Binary Search Tree: https://leetcode.com/problems/unique-binary-search-trees/ 这道题是一道BST的题,并且...
if(root->left) levelorder(root->left, level + 1, res); if(root->right) levelorder(root->right, level + 1, res); } }; 类似题目: [LeetCode] 107. Binary Tree Level Order Traversal II 二叉树层序遍历 II [LeetCode] 199. Binary Tree Right Side View 二叉树的右侧视图 All LeetCode Qu...
606. Construct String from Binary Tree 题解树的遍历 除递归方式遍历二叉树外,另可以借助堆栈(stack)实现二叉树中序、前序、后序遍历,使用队列(queue)实现按层遍历,例如 LeetCode题目 94. Binary Tree Inorder Traversal: // 94. Binary Tree Inorder Traversal vector<int> inorderTraversal(TreeNode* root)...
问题链接 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 the root of a binary tree, imagine yourself standing on the right side of it, return the values of the n…
1privatevoiddfsHelper(Map<Integer,Integer>depthToValue,TreeNode node,int depth){2if(node==null){3return;4}56// this is a pre-order traversal, essentially keep overwriting the depthToValue map (right view)7// while traverse the tree from left to right8depthToValue.put(depth,node.val);9...
【LeetCode】199. Binary Tree Right Side View 解题报告(Python) 标签: LeetCode 题目地址:https://leetcode.com/problems/binary-tree-right-side-view/description/ 题目描述: Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered...
TreeNode node = queue.poll(); if (temp == 0) { result.add(node.val); } if (node.left != null) { queue.addLast(node.left); len++; } if (node.right != null) { queue.addLast(node.right); len++; } } } return result; ...
private void pushAllTheLeft(Stack<TreeNode> s, TreeNode root){ s.push(root); while(root.left!=null){ root = root.left; s.push(root); } } } Binary Tree Postorder Traversal Given a binary tree, return the postorder traversal of its nodes' values. ...
{ public: TreeNode* invertTree(TreeNode* root) { fun(root); return root; } void fun(TreeNode* root) { if(root==NULL) return; if(root->left!=NULL) fun(root->left); if(root->right!=NULL) fun(root->right); TreeNode* term = root->left; root->left = root->right; root->...
0501-find-mode-in-binary-search-tree 0503-next-greater-element-ii 0505-the-maze-ii 0508-most-frequent-subtree-sum 0509-fibonacci-number 0513-find-bottom-left-tree-value 0515-find-largest-value-in-each-tree-row 0518-coin-change-ii 0523-continuous-subarray-sum 0525-contiguous-array 0526-beautiful...