111题

方法一：广度优先遍历（BFS）

/**
 * 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==NULL){
            return 0;
        }
        queue<pair<TreeNode*,int>>q;
        q.emplace(root,0);
        int result=0;
        while(!q.empty()){
            TreeNode* head=q.front().first;
            int dep=q.front().second;
            q.pop();
            if(head->left==NULL&&head->right==NULL)
            {
                 result=dep+1;
                 break;
            }
            if(head->left!=NULL){
                q.emplace(head->left,dep+1);
            }
            if(head->right!=NULL){
                q.emplace(head->right,dep+1);
            }
        }
        return result;
    }
};

方法二：深度遍历(DFS)

class Solution {
public:
    int minDepth(TreeNode* root) {
        if(root==NULL){
            return 0;
        }
        dfs(root,1);
        return result;
    }
    void dfs(TreeNode* root,int depth){
        if(root==NULL){
            return;
        }
        if(root->right==NULL&&root->left==NULL){
            if(result>depth){
                result=depth;
            }
        }  
        dfs(root->left,depth+1);   
        dfs(root->right,depth+1);
    }
    int result=INT_MAX;
};