力扣-111题二叉树的最小深度（C++）- 迭代+递归

怼烎@ 2022-09-02 11:57 144阅读 0赞

题目链接：[https://leetcode-cn.com/problems/minimum-depth-of-binary-tree/][https_leetcode-cn.com_problems_minimum-depth-of-binary-tree]  
题目如下：  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQwNDY3Njcw_size_16_color_FFFFFF_t_70]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQwNDY3Njcw_size_16_color_FFFFFF_t_70 1]  
解法一：迭代

/**
     * 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 minDepth(TreeNode* root) {
        
            int min_depth=0;
            if(root==NULL) return min_depth;
            queue<TreeNode*> que;
            que.push(root);
    
            while(!que.empty()){
        
                int size=que.size();
                min_depth++;
    
                while(size--){
        
                    TreeNode* node=que.front();
                    que.pop();
    
                    if(!node->left&&!node->right) return min_depth;
                    
                    if(node->left) que.push(node->left);
                    if(node->right) que.push(node->right);
                }
            }
    
            return min_depth;
        }
    };

解法二：递归

/**
     * 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* root){
        
        	//思路：如果左子树为空，右子树不为空，说明最小深度是 1 + 右子树的深度。
        	//反之，右子树为空，左子树不为空，最小深度是 1 + 左子树的深度。
        	//最后如果左右子树都不为空，返回左右子树深度最小值 + 1 。
            if(root==NULL) return 0;
            
            int left=dfs(root->left);//左
            int right=dfs(root->right);//右
            
            //中
            if(root->left==NULL&&root->right!=NULL) return right+1;
            if(root->left!=NULL&&root->right==NULL) return left+1;
            //if(root->left==NULL&&root->right==NULL) return left+1; 
    
            return min(left,right)+1;
        }
    
        int minDepth(TreeNode* root) {
        
            return dfs(root);
        }
    };

[https_leetcode-cn.com_problems_minimum-depth-of-binary-tree]: https://leetcode-cn.com/problems/minimum-depth-of-binary-tree/
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[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzQwNDY3Njcw_size_16_color_FFFFFF_t_70 1]: /images/20220829/d144002abf21458ba18545f04680ec6e.png