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79 lines
2.0 KiB
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79 lines
2.0 KiB
Plaintext
---
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title: '236. Lowest Common Ancestor of a Binary Tree'
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description: Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree
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icon: line-dot-left-horizontal
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sidebar:
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label: 'Lowest Common Ancestor of a Binary Tree'
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badge: 'Medium'
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---
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<Badge variant="accent">Tree</Badge>
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### Example 1:
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- Input: `root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1`
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- Output: `3`
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- Explanation: The LCA of nodes `5` and `1` is `3`.
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### Example 2:
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- Input: `root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4`
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- Output: `5`
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- Explanation: The LCA of nodes `5` and `4` is `5`, since a node can be a descendant of itself according to the LCA definition.
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### Example 3:
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- Input: `root = [1,2], p = 1, q = 2`
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- Output: `1`
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### Constraints:
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- The number of nodes in the tree is in the range [2, 10^5].
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- `-10^9 <= Node.val <= 10^9`
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- All Node.val are unique.
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- `p != q`
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- p and q will exist in the tree.
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## Approach
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```mermaid
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flowchart TD
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S(["lowestCommonAncestor(root, p, q)"]) --> B{"not root?"}
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B -- yes --> N(["return None"])
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B -- no --> M{"root == p or root == q?"}
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M -- yes --> R(["return root — a node is its own ancestor"])
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M -- no --> L["l = self.lowestCommonAncestor(root.left, p, q)"]
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L --> Q["r = self.lowestCommonAncestor(root.right, p, q)"]
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Q --> D{"l and r?"}
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D -- yes --> A(["return root — p and q were found on different sides"])
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D -- no --> E(["return l or r — pass up whichever side found something"])
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```
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## Solution
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```py
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def lowestCommonAncestor(
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self, root: "TreeNode", p: "TreeNode", q: "TreeNode"
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) -> "TreeNode":
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if not root:
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return None
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if root == p or root == q:
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return root
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l = self.lowestCommonAncestor(root.left, p, q)
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r = self.lowestCommonAncestor(root.right, p, q)
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if l and r:
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return root
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else:
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return l or r
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```
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