题目描述

从上到下按层打印二叉树，同一层的结点按从左到右的顺序打印，每一层打印到一行。

样例
输入如下图所示二叉树[8, 12, 2, null, null, 6, null, 4, null, null, null]
    8
   / \
  12  2
     /
    6
   /
  4

输出：[[8], [12, 2], [6], [4]]

算法1

(暴力枚举) $O(n)$

两个数组存上下两层的指针，两个数组存答案

C++ 代码

/**
 * 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>> printFromTopToBottom(TreeNode* root) {
        vector<vector<int>> res;

        if(!root) return res;
        vector<TreeNode*> first, second;
        first.push_back(root);
        int i = 0, j = 0;

        while(i < first.size()){
            auto t = first[i];
            ++i;
            if(t->left) second.push_back(t->left);
            if(t->right) second.push_back(t->right);

            if(i == first.size()){
                vector<int> tmp;
                for(int k = 0; k < first.size(); ++k){
                    auto node = first[k];
                    tmp.push_back(node->val);
                }
                res.push_back(tmp);
                tmp = {};

                i = 0;
                first = second;
                second = {};
            }
        }
        return res;
    }
};

算法2

(暴力枚举) $O(n)$

还是用队列，在每一层的后面加一个标记 nullptr///在root后边加个nullptr

C++ 代码

/**
 * 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>> printFromTopToBottom(TreeNode* root) {
        vector<vector<int>> res;
        if(!root) return res;
        queue<TreeNode*> q;
        q.push(root);
        q.push(nullptr);

        while(!q.empty() && q.front()){
            vector<int> tmp;

            while(q.front()){
                auto t = q.front();
                tmp.push_back(t->val);
                q.pop();

                if(t->left) q.push(t->left);
                if(t->right) q.push(t->right);
            }
            q.pop();

            q.push(nullptr);

            res.push_back(tmp);
        }
        return res;

    }
};