problem: 0236 LCA of a binary tree

Author: cychiu8

0236 Lowest Common Ancestor of a Binary Tree/0236 Lowest Common Ancestor of a Binary Tree.md

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+# 0236. Lowest Common Ancestor of a Binary Tree
+
+Difficulty: medium
+Link: https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/
+highlight: 紀錄是否在該個子樹內
+Topics: DFS-BFS
+
+# Clarification
+
+1. Check the inputs and outputs
+
+# Solution
+
+### Thought Process
+
+- Recursive [[ref-leetcode-solution]](https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/solution/)
+    - 條件
+        - current == null: return False
+        - else return mid or left or right
+            - mid = current == q or q
+            - left = dfs(root.left)
+            - right = dfs(root.right)
+    - 若 mid , left, right 有兩個 == True
+        - ⇒ 這個node 就是 LCA
+    
+    ![Untitled](./Untitled.png)
+    
+    ![Untitled](./Untitled%201.png)
+    
+    ![Untitled](./Untitled%202.png)
+    
+    ![Untitled](./Untitled%203.png)
+    
+    ![Untitled](./Untitled%204.png)
+    
+    ![Untitled](./Untitled%205.png)
+    
+    ![Untitled](./Untitled%206.png)
+    
+    ![Untitled](./Untitled%207.png)
+    
+    ![Untitled](./Untitled%208.png)
+    
+- Implement
+    
+    ```python
+    # Definition for a binary tree node.
+    # class TreeNode:
+    #     def __init__(self, x):
+    #         self.val = x
+    #         self.left = None
+    #         self.right = None
+    
+    class Solution:
+        def __init__(self):
+            self.ans = None
+        def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
+            
+            def dfs(current):
+                
+                if not current:
+                    return False
+                
+                left = dfs(current.left)
+                right = dfs(current.right)
+                mid = False
+                
+                if current == p or current == q:
+                    mid = True
+                
+                if mid + left + right >= 2:
+                    self.ans = current
+                
+                return mid or left or right
+            
+            dfs(root)
+            return self.ans
+    ```
+    
+
+### Complexity
+
+- Time complexity: O(N)
+    - N: number of nodes in binary tree
+    - search all nodes
+- Space complexity: O(N)
+    - recursion stack
+    - skewed binary tree