Leecode刷题之旅-C语言/python-111二叉树的最小深度

/*
 * @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。

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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