Minimum Depth of Binary Tree & Maximum 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.

方法一：后序遍历这颗树，先分别求出左右两颗子树的高度，在取最小，代码如下：

/**
* 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) {
if( !root ) return 0;   //空树，直接返回0
if( !root->left & !root->right ) return 1;  //如果是只有一个节点，则是1层
int left = minDepth(root->left);    //求左子树的高度
int right = minDepth(root->right);  //求右子树的高度
if( left == 0 ) return right + 1;   //如果左子树为空
if( right == 0 ) return left + 1;   //如果右子树为空
return min(left, right) + 1;    //返回左右子树中小的高度再加1
}
};

方法二：层次遍历这颗数，并记录当前节点所在的层，先访问到叶节点是，这时的层数就是最小的root到叶节点的路径，代码如下：

/**
* 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) {
if( !root ) return 0;
queue<TreeNode*> qt;
qt.push(root);
qt.push(NULL);
int level = 1;
while( !qt.empty() ) {
TreeNode* cur = qt.front();
qt.pop();
if( cur ) {
if( !cur->left && !cur->right ) break;  //如果当前节点就是叶节点，直接跳出循环
if( cur->left ) qt.push(cur->left);
if( cur->right ) qt.push(cur->right);
}
else {
if( !qt.empty() ) qt.push(NULL);    //如果队列不为空，则加入层标志
++level;    //进入下一层
}
}
return level;
}
};

Maximum Depth of Binary Tree也可以使用上述方法