Maximum Depth of Binary Tree

Given a binary tree, find its maximum depth.

Question

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

Example:

Input: tree = {}
Output: 0
Explanation: The height of empty tree is 0.


Input: tree = {1,2,3,#,#,4,5}
Output: 3	
Explanation: Like this:
   1
  / \                
 2   3                
    / \                
   4   5
it will be serialized {1,2,3,#,#,4,5}

Analyzation

Recursion

IMPORTANT PATTERN!! All binary tree problem that relates to “LEVEL” or “DEPTH” can use this pattern.

Import a deque package and use the queue structure. (Only 1 queue & BFS algorithm) See the code below.

The Code

from collections import deque
class Solution:
    """
    @param root: The root of binary tree.
    @return: An integer
    """
    def maxDepth(self, root):
        # write your code here
        if root is None:
            return 0
            
        queue = deque([root])
        result = []

        # while queue is not empty
        while queue:
            level = []
            for _ in range(len(queue)):
                node = queue.popleft() # get the first node of the queue
                level.append(node.val)
                
                # add children of the node to queue
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
                    
            # put the same level nodes in a list
            result.append(level)
        
        return len(result)

Time & Space Complexity

• Time complexity: O(n). Remember that the best case of a Binary Tree is O(logn) since it is a balanced tree but the worst case is O(n) because it can skew to only one side.
• Space complexity: O(n). Remember that the space complexity is always about how many nodes there are in a Binary Tree. If the tree is well-balanced then it is O(logn) but the worst case will be O(n).