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);}
};

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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;}
};