题目描述
给你二叉树的根节点 root ,返回其节点值的 锯齿形层序遍历 。(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行)。
示例 1:
输入:root = [3,9,20,null,null,15,7] 输出:[[3],[20,9],[15,7]]
示例 2:
输入:root = [1] 输出:[[1]]
个人C++解答
GYL
层序遍历模板,数组保存每一层,层的奇偶变化改变判断是偶逆置(reverse);
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: vector<vector<int>> zigzagLevelOrder(TreeNode* root) { if (root == nullptr) return {}; vector<TreeNode*> q; vector<vector<int>> ans; int flag = 0; q.push_back(root); while (!q.empty()) { vector<TreeNode*> vec; vector<int> temp; flag ++; for (int i = 0; i <q.size(); i++) { temp.push_back(q[i]->val); TreeNode* node = q[i]; if (node->left) vec.push_back(node->left); if (node->right) vec.push_back(node->right); } if(flag%2==0) reverse(temp.begin(),temp.end()); ans.push_back(temp); q = move(vec); } return ans; } };