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struct TreeNode {	int val;	TreeNode *left;	TreeNode *right;	TreeNode(int x) : val(x), left(NULL), right(NULL) {}};class Solution {	//find the minimum step of a tree, bfs works at most timepublic:	int BFS(TreeNode* root)	{		if(!root) 			return 0;        queue
   
    
     > q;		q.push(make_pair(root, 1));		while(!q.empty())		{			TreeNode* curNode = q.front().first;			int curStep = q.front().second;			q.pop();//note here			if(!curNode->left && !curNode->right)				return curStep;			if(curNode->left)				q.push(make_pair(curNode->left, curStep+1));			if(curNode->right)				q.push(make_pair(curNode->right, curStep+1));		}	}    int minDepth(TreeNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function		return BFS(root);    }};
    
   

second time

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int minDepth(TreeNode *root) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if(root == NULL) return 0;                if(root->left == NULL) return minDepth(root->right)+1;        else if(root->right == NULL) return minDepth(root->left)+1;        else return min(minDepth(root->left), minDepth(root->right))+1;    }};

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