題目：

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. Note: A leaf is a node with no children.

Example: Given binary tree [3,9,20,null,null,15,7],

    3   
    / \   
   9  20
     /  \    
    15   7 
 

return its minimum depth = 2.

解釋： 返回二叉樹的最小的深度。應該用dfs做，先求出左右兩個子樹的最小深度，如果左右子樹的深度中有一個是0或者兩個都是0，返回1+left+right（也就是1+深度不為0的子樹的深度），注意，一個葉子節點的深度是1。 python程式碼：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
 


class Solution(object):
    def minDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if not root:
            return 0
        left=self.minDepth(root.left)
        right=self.minDepth(root.right)
        if left==0 or right==0:
            return left+
 
right+1
        return 1+min(left,right)

c++程式碼：

/**
 * 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 0;
        int left=minDepth(root->left);
        int right =minDepth(root->right);
        if(!left ||!right)
            return 1+left+right;
        return 1+min(left,right);
    }
};

總結： 注意這裡和一般的dfs的解法不一樣，需要先進dfs，再對dfs之後的結果做判斷。