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111. Minimum Depth of Binary Tree 二叉树的最小深度

转载 作者:大佬之路 更新时间:2024-01-31 14:21:28 28 4
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题目地址:https://leetcode.com/problems/minimum-depth-of-binary-tree/open in new window

Total Accepted: 70767 Total Submissions: 243842 Difficulty: Easy

题目描述

Given a binary tree, find its minimum depth.

Theminimum 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.

题目大意

求根节点到最近的叶子节点的高度。

解题方法

DFS

运用递归,递归当前和 左子树和右子树的深度,某节点的左右子树都是空的时候,说明是叶子。计算根节点到此叶子的深度。

注意:如果是叶子,那么此叶子的深度是1.

同时注意:如果有一方的某一子树为空,那么它的深度为0,但不应该进入树的深度的计算当中去。

Better solution:用HashMap存储已经遍历过的树,减少空间复杂度。实现效率的提高。

1、 当root为空的时候直接返回0,因为MIN赋值很大,所以如果不单独预判的话会返回MIN;
2、 判断树的深度应该到叶子节点,也就是左右子结点都为空的那个结点;
3、 树的深度的根节点深度为1;

递归解法如下:

# 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
        que = collections.deque()
        que.append(root)
        depth = 1
        while que:
            size = len(que)
            for _ in range(size):
                node = que.popleft()
                if not node: continue
                if not node.left and not node.right:
                    return depth
                que.append(node.left)
                que.append(node.right)
            depth += 1
        return depth

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BFS

其实BFS更简单,因为发现这层有个叶子的话,就直接返回就行了。

# 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 not left:
            return right + 1
        if not right:
            return left + 1
        return 1 + min(left, right)

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