104. 二叉树的最大深度 - 力扣（LeetCode）

递归判断，当前节点的最大深度为1 + max(左节点的最大深度，右节点的最大深度)

/**
 * 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:
    int dfs(TreeNode *cur)
    {
        if (cur == nullptr) return 0;
        return 1 + max(dfs(cur->left), dfs(cur->right));
    }
    int maxDepth(TreeNode* root) {
        return dfs(root);
    }
};

102. 二叉树的层序遍历 - 力扣（LeetCode）

使用队列，每次出队前先记录当前队列的长度k，k为层的节点数量
此时只需要将前k个节点出队并将节点的左右子节点入队即可
在出队的同时维护答案

/**
 * 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>> ans;
        if (root == nullptr) return ans;
        queue<TreeNode*> q;
        q.push(root);
        while (q.size())
        {
            ans.push_back(vector<int>());
            int k = q.size();
            for (int i = 0; i < k; ++ i)
            {
                TreeNode* cur = q.front(); q.pop();
                ans.back().push_back(cur->val);
                if (cur->left) q.push(cur->left);
                if (cur->right) q.push(cur->right);
            }
        }
        return ans;
    }
};