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Leecode刷题之旅-C语言/python-111二叉树的最小深度
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/*
* @lc app=leetcode.cn id=111 lang=c
*
* [111] 二叉树的最小深度
*
* https://leetcode-cn.com/problems/minimum-depth-of-binary-tree/description/
*
* algorithms
* Easy (37.27%)
* Total Accepted: 12.2K
* Total Submissions: 32.6K
* Testcase Example: '[3,9,20,null,null,15,7]'
*
* 给定一个二叉树,找出其最小深度。
*
* 最小深度是从根节点到最近叶子节点的最短路径上的节点数量。
*
* 说明: 叶子节点是指没有子节点的节点。
*
* 示例:
*
* 给定二叉树 [3,9,20,null,null,15,7],
*
* 3
* / \
* 9 20
* / \
* 15 7
*
* 返回它的最小深度 2.
*
*/
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
int min(a,b);
int minDepth(struct TreeNode* root) {
if (root == NULL) return ;
if (root->left == NULL && root->right == NULL) return ;
if (root->left == NULL) return minDepth(root->right) + ;
else if (root->right == NULL) return minDepth(root->left) + ;
else return + min(minDepth(root->left), minDepth(root->right));
}
int min(a,b){
return a<b?a:b;
}
最小深度和最大深度类似,但是要注意的就是,当左子树为空的时候,只查右子树就可以,右子树为空的时候,只查左子树即可。
------------------------------------------------------------------------------------------------------------------------------------------------------------------------
python:
#
# @lc app=leetcode.cn id=111 lang=python3
#
# [111] 二叉树的最小深度
#
# https://leetcode-cn.com/problems/minimum-depth-of-binary-tree/description/
#
# algorithms
# Easy (37.27%)
# Total Accepted: 12.2K
# Total Submissions: 32.6K
# Testcase Example: '[3,9,20,null,null,15,7]'
#
# 给定一个二叉树,找出其最小深度。
#
# 最小深度是从根节点到最近叶子节点的最短路径上的节点数量。
#
# 说明: 叶子节点是指没有子节点的节点。
#
# 示例:
#
# 给定二叉树 [3,9,20,null,null,15,7],
#
# 3
# / \
# 9 20
# / \
# 15 7
#
# 返回它的最小深度 2.
#
#
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None # 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):
if root == None:
return 0
elif root.left == None and root.right == None:
return 1
elif root.left == None and root.right != None:
return self.minDepth(root.right)+1
elif root.right ==None and root.left != None:
return self.minDepth(root.left)+1
elif root.left != None and root.right !=None:
return min(self.minDepth(root.left), self.minDepth(root.right))+1






