LeetCode 1302 层数最深叶子节点的和[BFS DFS] HERODING的LeetCode之路

Posted 2022-08-18 HERODING23

解题思路：
对于二叉树中某一层节点的操作，DFS和BFS都可以轻松解决，DFS只要在递归时加入层次到达目的层进行累加即可，DFS就一层一层进行，直到最后一层，首先是DFS的代码：

/**
 * 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 
private:   
    int sum = 0;
    int depth = 0;
public:
    int deepestLeavesSum(TreeNode* root) 
        dfs(root, 0);
        return sum;
    

    void dfs(TreeNode* node, int layer) 
        if(node == nullptr) 
            return;
        
        if(layer > depth) 
            sum = node->val;
            depth = layer;
         else if(layer == depth) 
            sum += node->val;
        
        dfs(node->left, layer + 1);
        dfs(node->right, layer + 1);
    
;

BFS就是直接套用模板了，用队列存储每层的节点然后记录统计和，最后返回最后一层的和，代码如下：

/**
 * 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 deepestLeavesSum(TreeNode* root) 
        int sum;
        queue<TreeNode*> q;
        q.emplace(root);
        while(!q.empty()) 
            int n = q.size();
            sum = 0;
            for(int i = 0; i < n; i ++) 
                TreeNode* node = q.front();
                sum += node->val;
                q.pop();
                if(node->left != nullptr) 
                    q.emplace(node->left);
                
                if(node->right != nullptr) 
                    q.emplace(node->right);
                
            
        
        return sum;
    
;