Leetcode|二叉树的属性|111. 二叉树的最小深度

1 BFS（在最小深度问题上不用遍历所有树节点）

/**
 * 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) {
    
    
        if (!root) return 0;
        queue<TreeNode*> q;
        q.push(root);
        int depth = 1;
        while (!q.empty()) {
    
    
            int sz = q.size();
            for (int i=0;i < sz;i++) {
    
    
                TreeNode* e_ptr = q.front(); q.pop();
                if (!e_ptr->left && !e_ptr->right)
                    return depth;
                if (e_ptr->left) q.push(e_ptr->left);
                if (e_ptr->right) q.push(e_ptr->right);
            }
            depth++;
        }
        return depth;
    }
};

2 DFS（需要遍历所有树节点）

class Solution {
    
    
private:
    int min_depth = 1e4;
public:
    void dfs(TreeNode* root, int depth) {
    
    
        if (!root->left && !root->right && depth < min_depth)
            min_depth = depth;
        if (root->left)
            dfs(root->left, depth + 1);
        if (root->right)
            dfs(root->right, depth + 1);
    }
    int minDepth(TreeNode* root) {
    
    
        if (!root) return 0;
        dfs(root, 1);
        return min_depth;
    }
};