Leetcode 104. Maximum Depth of Binary Tree

Description: Given the root of a binary tree, return its maximum depth.

A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Link: 104. Maximum Depth of Binary Tree

Examples:

Example 1:

 

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

 Output: 3

Example 2:
Input: root = [1,null,2]
Output: 2

Example 3:
Input: root = []
Output: 0

Example 4:
Input: root = [0]
Output: 1

思路: 深度优先搜索所有路径，收集所有路径的深度，返回最大的深度。

class Solution(object):
    def maxDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        path = []
        self.dfs(root, 0, path)
        return max(path)
        
    def dfs(self, node, length, path):
        if node is None:
            path.append(length)
            return
        else:
            length += 1
        self.dfs(node.left, length, path)
        self.dfs(node.right, length, path)

但是很显然这样的方式浪费了不必要的存储空间，参考了一篇博客的DFS解题思路，很巧妙，很美。

class Solution(object):
    def maxDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        if root is None:
            return 0
        else:
            return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))

如何用BFS来做呢？BFS通常用队列来实现，遍历每一层的节点。

class Solution(object):
    def maxDepth(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        depth = 0
        que = collections.deque()   # 两端都可以插入删除的队列
        que.append(root)
        while que:                # not empty
            size = len(que)
            for _ in range(size):      # traverse all nodes of current layer, que just stores nodes of one layer
                node = que.popleft()
                if node is None:
                    continue
                que.append(node.left)
                que.append(node.right)
            depth += 1
        return depth-1    # take the tree only has root to know why depth-1

日期: 2021-03-12