1.题目描述 2.问题分析 使用层序遍历 3.代码 vector<int> v; vector<int> rightSideView(TreeNode* root) { if (root == NULL) return v; queue<TreeNode*> q; q.push(root); while (!q.empty()) { int size = q.size(); ; i < size; i++) { TreeNode *node = q.front()…

[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 no…

Binary Tree Right Side View Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. For example:Given the following binary tree, 1 <--- / \ 2 3 <--- \ \ 5 4 <-…

Binary Tree Right Side View Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. Example: Input: [1,2,3,null,5,null,4] Output: [1, 3, 4] Explanation: 1 <--- / \…

题目: Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. Note: Recursive solution is trivial, could you do it iteratively? 说明:1)下面有两种实现:递归(Recursive )与非递归(迭代iterati…

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. For example:Given the following binary tree, 1 <--- / \ 2 3 <--- \ \ 5 4 <--- You should return [1, 3,…

Given a binary tree, flatten it to a linked list in-place. For example,Given 1 / \ 2 5 / \ \ 3 4 6 The flattened tree should look like: 1 \ 2 \ 3 \ 4 \ 5 \ 6 Hints: If you notice carefully in the flattened tree, each node's right child points to the…

Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. For example:Given the following binary tree, 1 <--- / \ 2 3 <--- \ \ 5 4 <--- You should return [1, 3,…

题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 说明: 1)二叉树可空 2)思路:a.根据前序遍历的特点, 知前序序列(PreSequence)的首个元素(PreSequence[0])为二叉树的根(root),  然后在中序序列(InSequence)中查找此根(…

1.题目描述 2.问题分析 使用递归 3.代码 TreeNode* pruneTree(TreeNode* root) { if (root == NULL) return NULL; prun(root); return root; } void prun(TreeNode *root) { if (root == NULL) return; if (!has_ones(root->left)) root->left = NULL; if (!has_ones(root->right)…

1.题目描述 2.分析 找出最大元素,然后分割数组调用. 3.代码 TreeNode* constructMaximumBinaryTree(vector<int>& nums) { int size = nums.size(); ) return NULL; TreeNode *dummy = ); maxcont(dummy->left, , size-, nums); return dummy->left; } void maxcont(TreeNode* &…

1.题目描述 2.分析 使用递归. 3.代码 vector<string> ans; vector<string> binaryTreePaths(TreeNode* root) { if (root == NULL) return ans; leafPath(root,""); return ans; } void leafPath(TreeNode *root, string s) { if (root == NULL) return ; if (root-…

1.题目描述 2.题目分析 先遍历,再反转. 3.代码 vector<vector<int>> levelOrderBottom(TreeNode* root) { vector<vector<int>> ans; if (root == NULL) return ans; queue<TreeNode*> q; q.push(root); vector<int> v; while (!q.empty()) { int size =…

1.题目描述 2.分析 利用递归实现. 3.代码 int findTilt(TreeNode* root) { if (root == NULL) ; ; nodesTilt(root,ans); return ans; } int nodesTilt(TreeNode *root, int & ans) { if (root == NULL) ; int tiltleft = nodesTilt(root->left, ans); int tiltright = nodesTilt(roo…

1.题目描述 2.问题分析 递归 3.代码 vector<int> postorderTraversal(TreeNode* root) { vector<int> v; postBorder(root,v); return v; } void postBorder(TreeNode *root, vector<int> &v) { if (root == NULL) return ; postBorder(root->left, v); postBord…

1.题目描述 2.问题分析 利用递归. 3.代码 vector<int> preorderTraversal(TreeNode* root) { vector<int> v; preBorder(root, v); return v; } void preBorder(TreeNode *root, vector<int> &v) { if (root == NULL) return ; v.push_back(root->val); preBorder(…

// 我的代码 package Leetcode; /** * 199. Binary Tree Right Side View * address: https://leetcode.com/problems/binary-tree-right-side-view/ * Given a binary tree, imagine yourself standing on the right side of it, * return the values of the nodes you can…

leetcode 199. Binary Tree Right Side View 这个题实际上就是把每一行最右侧的树打印出来,所以实际上还是一个层次遍历. 依旧利用之前层次遍历的代码,每次大的循环存储的是一行的节点,最后一个节点就是想要的那个节点 class Solution { public: vector<int> rightSideView(TreeNode* root) { vector<int> result; if(root == NULL) return resul…

Leetcode之深度优先搜索(DFS)专题-199. 二叉树的右视图(Binary Tree Right Side View) 深度优先搜索的解题详细介绍,点击 给定一棵二叉树,想象自己站在它的右侧,按照从顶部到底部的顺序,返回从右侧所能看到的节点值. 示例: 输入: [1,2,3,null,5,null,4] 输出: [1, 3, 4] 解释: 1 <--- / \ 2 3 <--- \ \ 5 4 <--- 分析:把每一行的值保存起来,最后再把每一行最后一个放进ans里. AC代码…

199. 二叉树的右视图 199. Binary Tree Right Side View 题目描述 给定一棵二叉树,想象自己站在它的右侧,按照从顶部到底部的顺序,返回从右侧所能看到的节点值. Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom. LeetCode19…

剑指offer 面试题39:判断平衡二叉树 提交网址:  http://www.nowcoder.com/practice/8b3b95850edb4115918ecebdf1b4d222?tpId=13&tqId=11192 时间限制:1秒       空间限制:32768K      参与人数:2481 题目描述 输入一棵二叉树,判断该二叉树是否是平衡二叉树. 分析: 平衡二叉树定义 递归解法 AC代码: #include<iostream> #include<vector&…

Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes. Left boundary is defined as the path from root to the…

Problem Link: http://oj.leetcode.com/problems/binary-tree-postorder-traversal/ The post-order-traversal of a binary tree is a classic problem, the recursive way to solve it is really straightforward, the pseudo-code is as follows. RECURSIVE-POST-ORDE…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 来源: https://leetcode.com/problems/binary-tree-level-order-traversal-ii/ Given a binary tree, return the bottom-up level order traversal of its nodes' values.…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 来源: https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/ Given a binary tree, return the zigzag level order traversal of its nodes' values…

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 = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 / \ 9 2…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 / \ 9 2…

LeetCode--Diameter of Binary Tree Question Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 / \ 9 2…