LeetCode102. Binary Tree Level Order Traversal

一、题目

Given the root of a binary tree, return the level order traversal of its nodes’ values. (i.e., from left to right, level by level).

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: [[3],[9,20],[15,7]]
Example 2:

Input: root = [1]
Output: [[1]]
Example 3:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 2000].
-1000 <= Node.val <= 1000

二、题解

/**
 * 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) {
        vector<vector<int>> res;
        queue<TreeNode*> q;
        if(root != nullptr) q.push(root);
        while(!q.empty()){
            int size = q.size();
            //记录本层元素
            vector<int> tmp;
            while(size--){
                TreeNode* t = q.front();
                q.pop();
                tmp.push_back(t->val);
                if(t->left != nullptr) q.push(t->left);
                if(t->right != nullptr) q.push(t->right);
            }
            res.push_back(tmp);
        }
        return res;
    }
};