111.minimum-depth-of-binary-tree

Statement

简体中文English

Metadata

• Link: 二叉树的最小深度
• Difficulty: Easy
• Tag: 树 深度优先搜索 广度优先搜索 二叉树

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 1：

输入：root = [3,9,20,null,null,15,7]
输出：2

示例 2：

输入：root = [2,null,3,null,4,null,5,null,6]
输出：5

 

提示：

• 树中节点数的范围在 [0, 105] 内
• -1000 <= Node.val <= 1000

Metadata

• Link: Minimum Depth of Binary Tree
• Difficulty: Easy
• Tag: Tree Depth-First Search Breadth-First Search 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.

Note: A leaf is a node with no children.

 

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 2

Example 2:

Input: root = [2,null,3,null,4,null,5,null,6]
Output: 5

 

Constraints:

• The number of nodes in the tree is in the range [0, 105].
• -1000 <= Node.val <= 1000

Solution

Python 3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def __init__(self):
        self.res = 1000000

    def dfs(self, rt: TreeNode, d: int) -> None:
        if not rt.left and not rt.right:
            self.res = min(self.res, d)

        if rt.left:
            self.dfs(rt.left, d + 1)

        if rt.right:
            self.dfs(rt.right, d + 1)

    def minDepth(self, root: TreeNode) -> int:
        if not root:
            return 0

        self.dfs(root, 1)
        return self.res

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最后更新: October 11, 2023