从上往下打印二叉树这个是不分行的,用一个队列就可以实现 class Solution { public: vector<int> PrintFromTopToBottom(TreeNode* root) { vector<int> result; queue<TreeNode* > container; if(root == NULL) return result; container.push(root); ){ TreeNode* rot = container.f…

题目: Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3], [9,20], [15,7]…

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3], [9,20], [15,7…

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3], [9,20], [15,7…

102. 二叉树的层次遍历 (1过,隐蔽错误花时间很多,简单题目本应很快,下次注意红色错误的地方) 给定一个二叉树,返回其按层次遍历的节点值. (即逐层地,从左到右访问所有节点). 例如:给定二叉树: [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 返回其层次遍历结果: [ [3], [9,20], [15,7] ] public static ArrayList<ArrayList<Integer>> levelOrder(TreeNod…

本题是广度优先遍历(BFS)实现树的层次遍历,使用队列实现. class Solution { public: vector<vector<int>> levelOrder(TreeNode* root) { vector<vector<int>> res; queue<TreeNode*>q; if (root != NULL) { q.push(root); while (!q.empty()) { vector<int> tmp…

题目 给定一个二叉树,返回其按层次遍历的节点值. (即逐层地,从左到右访问所有节点). 例如: 给定二叉树: [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 返回其层次遍历结果: [ [3], [9,20], [15,7] ] 考点 思路 代码 newcoder /* struct TreeNode { int val; struct TreeNode *left; struct TreeNode *right; TreeNode(int x) : va…

题面 Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). 层序遍历二叉树,要求从上到下,从左到右,输出结果为二维vector. 样例 Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its level order trave…

102. 二叉树的层次遍历 描述 给定一个二叉树,返回其按层次遍历的节点值. (即逐层地,从左到右访问所有节点). 示例 例如,给定二叉树: [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 返回其层次遍历结果: [ [3], [9,20], [15,7] ] 思路 关于树的问题基本上都是用递归来解决的,此题也是需要考虑递归. 对树,常使用的查找方法有深度优先查找和广度优先查找,在此使用 DFS ,因为它的代码易于理解. 层次遍历的核心在于需要设置一个层数…

给定一个二叉树,返回其按层次遍历的节点值. (即逐层地,从左到右访问所有节点). 例如: 给定二叉树: [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 返回其层次遍历结果: [ [3], [9,20], [15,7] ] class Solution { public: vector<vector<int> > levelOrder(TreeNode* root) { vector<vector<int> > res…