网址:https://leetcode.com/problems/binary-tree-level-order-traversal/
参考:https://www.cnblogs.com/grandyang/p/4051321.html
二叉树的层次遍历的做法是维护一个队列,不断使节点入队,使用了val后即出队。结束标志是队列是否为空
主要难度在于要分别提取出每层的val
我们可把在树的一层节点全部入列后队列的长度作为for循环的判断依据
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int>> levelOrder(TreeNode* root) { if (!root) return {}; vector<vector<int>> res; queue<TreeNode*> q; q.push(root); vector<int> oneLevel; while (!q.empty()) { oneLevel.clear(); // 一轮for循环即是对一层节点的处理 for (int i = q.size(); i > 0; --i) { TreeNode *t = q.front(); q.pop(); oneLevel.push_back(t->val); if (t->left) q.push(t->left); if (t->right) q.push(t->right); } res.push_back(oneLevel); } return res; } };