111. Minimum Depth of Binary Tree
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:
mine思路:
1.采用后序遍历非递归方式,找到叶子节点,记录当时栈内元素的个数到容器中。
2.最后从容器中选择最小的一个输出。
代码如下:
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/** * 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) {
int iMinDepth;
vector< int > depths;
stack<TreeNode *> s;
TreeNode *p,*q;
q = NULL;
p = root;
if (!root)
return 0;
while (p != NULL || s.size() > 0)
{
while ( p != NULL)
{
s.push(p);
p = p->left;
}
if (s.size() > 0)
{
p = s.top();
if ( NULL == p->left && NULL == p->right)
{
depths.push_back(s.size());
}
if ( (NULL == p->right || p->right == q) )
{
q = p;
s.pop();
p = NULL;
}
else
p = p->right;
}
}
iMinDepth = depths[0];
for ( int i = 0; i < depths.size(); i++)
{
if (depths[i] < iMinDepth)
{
iMinDepth = depths[i];
}
}
return iMinDepth;
}
}; |
参考其他:http://www.cnblogs.com/bakari/p/4126693.html
有两种求解的思路,一种采用DFS的思想,一种采用BFS的思想,如下代码所示:
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struct TreeNode {
int val;
TreeNode* left;
TreeNode* right;
TreeNode( int x): val(x), left(NULL),right(NULL) {}
}; //采用DFS的思想 int maxDepth(TreeNode *root)
{ if (NULL == root)
return 0;
int l = maxDepth(root->left);
int r = maxDepth(root->right);
return l > r ? l + 1:r+1;
//以上这两种方式有一种更简便的方法
//return 1 + max(maxDepth(root->left), maxDepth(root->right));
} //采用BFS的方法,引入队列 int maxDepth(TreeNode *root)
{ if (NULL == root)
return 0;
queue <TreeNode *> que;
int nCount = 1;
int nDepth = 0; // 记录队列里面每一层上的元素
que.push(root);
while (!que.empty()) {
TreeNode *pTemp = que.front();
que.pop();
nCount --;
if (pTemp->left)
que.push(pTemp->left);
if (pTemp->right)
que.push(pTemp->right);
if (nCount == 0) {
nDepth ++;
nCount = que.size();
}
}
return nDepth;
} |
本文转自313119992 51CTO博客,原文链接:http://blog.51cto.com/qiaopeng688/1835237