Leetcode102. 二叉树的层序遍历——BFS

法一：层数和节点一起入队

/**
 * 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<pair<TreeNode*, int>> q;
        vector<vector<int>> res;

        if(root != NULL) q.push({root, 0});

        while(!q.empty()){
            TreeNode* node = q.front().first;
            int level = q.front().second;
            if(node -> left) q.push({node -> left, level + 1});
            if(node -> right) q.push({node -> right, level + 1});
            q.pop();
            if(res.size() < level + 1) res.push_back({}); //拓展一行
            res[level].emplace_back(node -> val);
        }

        return res;
    }
};

法二：在每一层遍历开始前，先记录队列中的结点数量 （也就是这一层的结点数量），然后一口气处理完这一层的 n 个结点。

/**
 * 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*> q;
        vector<vector<int>> res;

        if(root != NULL) q.push(root);

        while(!q.empty()){
            int size = q.size();
            vector<int> vec;
            for(int i = 0; i < size; ++i){
                vec.emplace_back(q.front() -> val);
                if(q.front() -> left) q.push(q.front() -> left);
                if(q.front() -> right) q.push(q.front() -> right);
                q.pop();
            }
            res.emplace_back(vec);
        }

        return res;
    }
};

学习到的语法点：

• 创建一个空的vector：res.push_back(vector ()) 或 res.push_back({})