# LeetCode(111)Minimum Depth of Binary Tree

## 题目描述

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.

• 左右子树都不为空
• 左子树为空，右子树不为空
• 左子树不为空，右子树为空

## C++代码

// 90ms过大集合
/** * 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) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
if(root!=NULL&&root->left!=NULL&&root->right!=NULL){
int l=minDepth(root->left);
int r=minDepth(root->right);
return (l<r?l:r)+1;
}else if(root!=NULL&&root->left==NULL){
return 1+minDepth(root->right);
}else if(root!=NULL&&root->right==NULL){
return 1+minDepth(root->left);
}else{
return 0;
}
}
};

## Python3代码

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

class Solution:
def minDepth(self, root):
""" :type root: TreeNode :rtype: int """
if root!=None and root.left!=None and root.right!=None:
ld = self.minDepth(root.left)
rd = self.minDepth(root.right)
return 1+(ld if ld < rd else rd)
elif root!=None and root.left == None:
return 1+self.minDepth(root.right)
elif root!=None and root.right == None:
return 1+self.minDepth(root.left)
else:
return 0