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feat: refactored docs structure
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-55
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---
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title: '104. Maximum Depth of Binary Tree'
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description: Given the root of a binary tree, return its maximum depth
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icon: dot
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topics:
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- { name: "Tree", slug: "tree" }
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- { name: "Depth-First Search", slug: "depth-first-search" }
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- { name: "Breadth-First Search", slug: "breadth-first-search" }
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- { name: "Binary Tree", slug: "binary-tree" }
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video: "hTM3phVI6YQ"
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issue: "https://github.com/prdlk/leetcode/issues/97"
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sidebar:
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label: 'Maximum Depth of Binary Tree'
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badge: 'Easy'
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---
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### Example 1:
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- Input: `root = [3,9,20,null,null,15,7]`
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- Output: `3`
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### Example 2:
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- Input: `root = [1,null,2]`
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- Output: `2`
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### Constraints:
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- The number of nodes in the tree is in the range [0, 10^4].
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- `-100 <= Node.val <= 100`
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## Approach
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```mermaid
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flowchart TD
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S(["maxDepth(root)"]) --> B{"root is None?"}
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B -- yes --> Z(["return 0"])
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B -- no --> L["l = maxDepth(root.left), r = maxDepth(root.right)"]
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L --> E(["return 1 + max(l, r) — this node adds one level above its deeper subtree"])
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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, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def maxDepth(self, root: Optional[TreeNode]) -> int:
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if not root:
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return 0
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return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))
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```
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---
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title: '226. Invert Binary Tree'
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description: Given the root of a binary tree, invert the tree, and return its root
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icon: dot
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topics:
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- { name: "Tree", slug: "tree" }
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- { name: "Depth-First Search", slug: "depth-first-search" }
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- { name: "Breadth-First Search", slug: "breadth-first-search" }
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- { name: "Binary Tree", slug: "binary-tree" }
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video: "OnSn2XEQ4MY"
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issue: "https://github.com/prdlk/leetcode/issues/96"
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sidebar:
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label: 'Invert Binary Tree'
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badge: 'Easy'
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---
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### Example 1:
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- Input: `root = [4,2,7,1,3,6,9]`
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- Output: `[4,7,2,9,6,3,1]`
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### Example 2:
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- Input: `root = [2,1,3]`
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- Output: `[2,3,1]`
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### Example 3:
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- Input: `root = []`
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- Output: `[]`
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### Constraints:
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- The number of nodes in the tree is in the range [0, 100].
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- `-100 <= Node.val <= 100`
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## Approach
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```mermaid
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flowchart TD
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S(["invertTree(root)"]) --> B{"root is None?"}
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B -- yes --> Z(["return None"])
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B -- no --> L["left = invertTree(root.left), right = invertTree(root.right) — both subtrees are already inverted"]
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L --> X["root.left = right, root.right = left"]
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X --> E(["return root"])
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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, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
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if not root:
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return None
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left = self.invertTree(root.left)
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right = self.invertTree(root.right)
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root.left = right
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root.right = left
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return root
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```
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-83
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---
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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: dot
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topics:
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- { name: "Tree", slug: "tree" }
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- { name: "Depth-First Search", slug: "depth-first-search" }
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- { name: "Binary Tree", slug: "binary-tree" }
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- { name: "Binary Lifting", slug: "binary-lifting" }
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- { name: "Lowest Common Ancestor", slug: "lowest-common-ancestor" }
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issue: "https://github.com/prdlk/leetcode/issues/98"
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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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### 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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@@ -1,5 +0,0 @@
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---
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title: Tree DFS
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description: 'Each branch is followed to its end before the next branch is examined.'
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hidden: true
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---
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import { defineMeta } from "blume";
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export default defineMeta({ title: "Tree DFS", order: 11, icon: "folder-tree", display: "page" });
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