思路:简单BFS即可
/** * 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>> levelOrder(TreeNode* root) { queue<TreeNode*> que; if (root != NULL) que.push(root); vector<vector<int>> result; while (!que.empty()) { int size = que.size(); vector<int> vec; for (int i = 0; i < size; i++) { TreeNode* node = que.front(); que.pop(); vec.push_back(node->val); if (node->left) que.push(node->left); if (node->right) que.push(node->right); } result.push_back(vec); } return result; } };