1--二叉树的层序遍历（102）

主要思路：

经典广度优先搜索，基于队列；

对于本题需要将同一层的节点放在一个数组中，因此遍历的时候需要用一个变量 nums 来记录当前层的节点数，即 nums 等于队列元素的数目；

#include <iostream>
#include <vector>
#include <queue>
 
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:
    std::vector<std::vector<int>> levelOrder(TreeNode* root) {
        std::vector<std::vector<int>> res;
        if(root == nullptr) return res;
        std::queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()){
            int nums = q.size(); // 当前层的节点数
            std::vector<int> tmp;
            while(nums > 0){ // 遍历处理同一层
                TreeNode *cur = q.front();
                q.pop();
                tmp.push_back(cur->val);
 
                if(cur->left != nullptr) q.push(cur->left);
                if(cur->right != nullptr) q.push(cur->right);
                
                nums--;
            }
            res.push_back(tmp); // 记录当前层的元素
        }
        return res;
    }
};
 
int main(int argc, char* argv[]){
    // root = [1, null, 2, 3]
    TreeNode *Node1 = new TreeNode(3);
    TreeNode *Node2 = new TreeNode(9);
    TreeNode *Node3 = new TreeNode(20);
    TreeNode *Node4 = new TreeNode(15);
    TreeNode *Node5 = new TreeNode(7);
    Node1->left = Node2;
    Node1->right = Node3;
    Node3->left = Node4;
    Node3->right = Node5;
 
    Solution S1;
    std::vector<std::vector<int>> res = S1.levelOrder(Node1);
    for(auto item : res) {
        for (int v : item) std::cout << v << " ";
        std::cout << std::endl;
    }
    return 0;
}

2--二叉树的最大深度

主要思路：

递归计算左右子树的深度，选取两者最大值 +1 返回；

#include <iostream>
#include <vector>
#include <queue>
 
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 maxDepth(TreeNode* root) {
        if(root == nullptr) return 0;
        int res = dfs(root);
        return res;
    }
 
    int dfs(TreeNode* root){
        if(root == nullptr) return 0;
        int left_height = dfs(root->left);
        int right_height = dfs(root->right);
        int cur_height = std::max(left_height, right_height) + 1;
        return cur_height;
    }
};
 
int main(int argc, char* argv[]){
    // root = [3,9,20,null,null,15,7]
    TreeNode *Node1 = new TreeNode(3);
    TreeNode *Node2 = new TreeNode(9);
    TreeNode *Node3 = new TreeNode(20);
    TreeNode *Node4 = new TreeNode(15);
    TreeNode *Node5 = new TreeNode(7);
 
    Node1->left = Node2;
    Node1->right = Node3;
    Node3->left = Node4;
    Node3->right = Node5;
 
    Solution S1;
    int res = S1.maxDepth(Node1);
    std::cout << res << std::endl;
    return 0;
}