leetcode之Minimum Depth of Binary Tree
2016-01-23 19:35
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题目:
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.
解答:
很简单的DFS,可以加入一个剪枝
/**
* 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:
void DFS(TreeNode *root,int cur)
{
if(cur + 1> minDep)
return;
if(root->left == NULL && root->right == NULL)
{
if(cur + 1 < minDep)
minDep = cur + 1;
return;
}
if(root->left != NULL)
DFS(root->left,cur + 1);
if(root->right != NULL)
DFS(root->right, cur + 1);
}
int minDepth(TreeNode* root) {
minDep = 1 << 30;
if(root == NULL)
return 0;
DFS(root,0);
return minDep;
}
private:
int minDep;
};
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.
解答:
很简单的DFS,可以加入一个剪枝
/**
* 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:
void DFS(TreeNode *root,int cur)
{
if(cur + 1> minDep)
return;
if(root->left == NULL && root->right == NULL)
{
if(cur + 1 < minDep)
minDep = cur + 1;
return;
}
if(root->left != NULL)
DFS(root->left,cur + 1);
if(root->right != NULL)
DFS(root->right, cur + 1);
}
int minDepth(TreeNode* root) {
minDep = 1 << 30;
if(root == NULL)
return 0;
DFS(root,0);
return minDep;
}
private:
int minDep;
};
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