solve(w04): 104. Maximum Depth of Binary Tree

Author: seungriyou

maximum-depth-of-binary-tree/seungriyou.py

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+# https://leetcode.com/problems/maximum-depth-of-binary-tree/
+
+from typing import Optional
+
+# 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 maxDepth_recur(self, root: Optional[TreeNode]) -> int:
+        """
+        [Complexity]
+            - TC: O(n) (모든 node를 한 번씩 방문)
+            - SC: O(h) (call stack) (h = logn ~ n)
+
+        [Approach] recursive
+            재귀적으로 max(left subtree의 depth, right subtree의 depth) + 1 을 구하면 된다.
+            base condition(= 현재 노드가 None인 경우)에서는 0을 반환한다.
+        """
+
+        def get_max_depth(node):
+            # base condition
+            if not node:
+                return 0
+
+            # recur
+            return max(get_max_depth(node.right), get_max_depth(node.left)) + 1
+
+        return get_max_depth(root)
+
+    def maxDepth(self, root: Optional[TreeNode]) -> int:
+        """
+        [Complexity]
+            - TC: O(n) (모든 node를 한 번씩 방문)
+            - SC: O(w) (트리의 너비) (w = 1 ~ n / 2)
+
+        [Approach] iterative
+            BFS 처럼 이진 트리를 레벨 별로 순회하며 depth를 1씩 증가시킬 수 있다.
+        """
+        depth = 0
+        curr_level = [root] if root else []
+
+        while curr_level:
+            next_level = []
+            for node in curr_level:
+                if node.left:
+                    next_level.append(node.left)
+                if node.right:
+                    next_level.append(node.right)
+
+            curr_level = next_level
+            depth += 1
+
+        return depth