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.

代码:

/**
* 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) return ;
if ( !root->left && !root->right) return ;
if ( root->left && root->right ) return std::min( Solution::minDepth(root->left)+, Solution::minDepth(root->right)+);
if ( root->left ) return Solution::minDepth(root->left)+;
if (root->right ) return Solution::minDepth(root->right)+;
}
};

tips:

深搜思路(递归实现)。这里需要控制向下进行的条件。

1. 如果root是NULL,返回0

2. 如果root不是NULL,且left和right都是NULL,则到达叶子节点返回1(代表算上叶子节点的那一层)

3. 如果root->left和root->right都不为NULL,则继续往两边深搜

4. 如果root不是NULL,但root->left或root->right哪一方为NULL,则为NULL的一端不会再有叶子节点出现,不能再往下走了。

完毕。

============================================

第二次过这道题,上来没有把终止条件想完全。终止条件应该是达到叶子简单root->left root->right都为空。

修改了一次后AC了。

/**
* 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 ) return ;
int min_depth = INT_MAX;
Solution::depth(root, min_depth, );
return min_depth;
}
static void depth(TreeNode* root, int& min_depth, int dep)
{
if ( !root->left && !root->right ) min_depth = min(min_depth,dep);
if ( root->left ) Solution::depth(root->left, min_depth, dep+);
if ( root->right ) Solution::depth(root->right, min_depth, dep+);
}
};

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