【题目描述】
Given a binary tree, find its minimum depth.
The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
【思路】一般这样的题目既可以用递归的方法也可以用非递归迭代的方法做。我是用递归的方法做的,思路基本和Maximum Depth of Binary Tree的类似。
【代码】
递归:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int minDepth(TreeNode* root) {
if(root==NULL) return 0;
if(!root->left&&!root->right) return 1;
if(root->left==NULL) return minDepth(root->right)+1;
else if(root->right==NULL) return minDepth(root->left)+1;
return min(minDepth(root->right),minDepth(root->left))+1;
}
};
//参考了soulmachine的思路
//pair是一种模板类型,包含两个数据值,两个数据的类型可以不同,pair<typename1,typename2> a,访问的话用a.first,a.second访问
//make_pair函数,自动生成pair对象
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int minDepth(TreeNode* root) {
if(root==NULL) return 0;
int result=INT_MAX;
stack<pair<TreeNode*,int>> s;
s.push(make_pair(root,1));
while(!s.empty()){
TreeNode* node=s.top().first ;
int val=s.top().second ;
s.pop();
if(node->left==NULL&&node->right==NULL) result=min(result,val);
if(node->left&&result>val) s.push(make_pair(node->left,val+1));
if(node->right&&result>val) s.push(make_pair(node->right,val+1));
}
return result;
}
};