Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 retur…

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 retur…

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 retur…

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example:Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 retur…

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its…

Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). For example: Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its…

原题地址 基本数据结构操作,二叉树的层次遍历. 代码: vector<vector<int> > zigzagLevelOrder(TreeNode *root) { vector<vector<int> > res; vector<TreeNode *> layer; bool l2r = true; layer.push_back(root); while (!layer.empty()) { vector<TreeNode *>…

原题链接 题目要求以"Z"字型遍历二叉树,并存储在二维数组里. 利用BFS,对每一层进行遍历.对于每一层是从左还是从右,用一个整数型判断当前是偶数行还是奇数行就可以了. class Solution { public: vector<vector<int>> res; vector<vector<int>> zigzagLevelOrder(TreeNode *root) { if (root == NULL) return res; s…

相对于102题,稍微改变下方法就行 迭代方法: 在102题的基础上,加上一个变量来判断是不是需要反转 反转的话,当前list在for循环结束后用collection的反转方法就可以实现反转 递归方法: 由于有层数,所以用层数%2判断是不是需要反转 反转的话就元素都添加到最前边,一层添加完后就是反的 下边是递归方法 List<List<Integer>> res = new ArrayList<>(); public List<List<Integer>…

// 103. Binary Tree Zigzag Level Order Traversal // https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/description/ // swift // Definition for a binary tree node. public class TreeNode { public var val: Int public var left: TreeNo…