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---
title: '104. Maximum Depth of Binary Tree'
description: Given the root of a binary tree, return its maximum depth
sidebar:
label: 'Maximum Depth of Binary Tree'
badge: 'Easy'
---
<Badge variant="accent">Tree DFS</Badge>
### Example 1:
- Input: `root = [3,9,20,null,null,15,7]`
- Output: `3`
### Example 2:
- Input: `root = [1,null,2]`
- Output: `2`
### Constraints:
- The number of nodes in the tree is in the range [0, 10^4].
- `-100 <= Node.val <= 100`
## Solution
```py
# 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(self, root: Optional[TreeNode]) -> int:
if not root:
return 0
return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))
```
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---
title: '226. Invert Binary Tree'
description: Given the root of a binary tree, invert the tree, and return its root
sidebar:
label: 'Invert Binary Tree'
badge: 'Easy'
---
<Badge variant="accent">Tree DFS</Badge>
### Example 1:
- Input: `root = [4,2,7,1,3,6,9]`
- Output: `[4,7,2,9,6,3,1]`
### Example 2:
- Input: `root = [2,1,3]`
- Output: `[2,3,1]`
### Example 3:
- Input: `root = []`
- Output: `[]`
### Constraints:
- The number of nodes in the tree is in the range [0, 100].
- `-100 <= Node.val <= 100`
## Solution
```py
# 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
if not root:
return None
left = self.invertTree(root.left)
right = self.invertTree(root.right)
root.left = right
root.right = left
return root
```
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---
title: '236. Lowest Common Ancestor of a Binary Tree'
description: Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree
sidebar:
label: 'Lowest Common Ancestor of a Binary Tree'
badge: 'Medium'
---
<Badge variant="accent">Tree DFS</Badge>
### Example 1:
- Input: `root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1`
- Output: `3`
- Explanation: The LCA of nodes `5` and `1` is `3`.
### Example 2:
- Input: `root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4`
- Output: `5`
- Explanation: The LCA of nodes `5` and `4` is `5`, since a node can be a descendant of itself according to the LCA definition.
### Example 3:
- Input: `root = [1,2], p = 1, q = 2`
- Output: `1`
### Constraints:
- The number of nodes in the tree is in the range [2, 10^5].
- `-10^9 <= Node.val <= 10^9`
- All Node.val are unique.
- `p != q`
- p and q will exist in the tree.
## Solution
```py
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def lowestCommonAncestor(
self, root: "TreeNode", p: "TreeNode", q: "TreeNode"
) -> "TreeNode":
if not root:
return None
if root == p or root == q:
return root
l = self.lowestCommonAncestor(root.left, p, q)
r = self.lowestCommonAncestor(root.right, p, q)
if l and r:
return root
else:
return l or r
```
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"""
104. Maximum Depth of Binary Tree
Difficulty: Easy
https://leetcode.com/problems/maximum-depth-of-binary-tree/
──────────────────────────────────────────────────
Given the root of a binary tree, return its maximum depth.
A binary tree's maximum depth is the number of nodes along the
longest path from the root node down to the farthest leaf node.
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: 3
Example 2:
Input: root = [1,null,2]
Output: 2
Constraints:
• The number of nodes in the tree is in the range [0, 10^4].
• -100 <= Node.val <= 100
"""
# 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(self, root: Optional[TreeNode]) -> int:
if not root:
return 0
return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))
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"""
226. Invert Binary Tree
Difficulty: Easy
https://leetcode.com/problems/invert-binary-tree/
──────────────────────────────────────────────────
Given the root of a binary tree, invert the tree, and return its root.
Example 1:
Input: root = [4,2,7,1,3,6,9]
Output: [4,7,2,9,6,3,1]
Example 2:
Input: root = [2,1,3]
Output: [2,3,1]
Example 3:
Input: root = []
Output: []
Constraints:
• The number of nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100
"""
# 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 invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
if not root:
return None
left = self.invertTree(root.left)
right = self.invertTree(root.right)
root.left = right
root.right = left
return root
@@ -0,0 +1,76 @@
"""
236. Lowest Common Ancestor of a Binary Tree
Difficulty: Medium
https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/
──────────────────────────────────────────────────
Given a binary tree, find the lowest common ancestor (LCA) of two
given nodes in the tree.
According to the definition of LCA on Wikipedia: &ldquo;The lowest
common ancestor is defined between two nodes p and q as the lowest
node in T that has both p and q as descendants (where we allow a node
to be a descendant of itself).&rdquo;
Example 1:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.
Example 2:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a
descendant of itself according to the LCA definition.
Example 3:
Input: root = [1,2], p = 1, q = 2
Output: 1
Constraints:
• The number of nodes in the tree is in the range [2, 10^5].
• -10^9 <= Node.val <= 10^9
• All Node.val are unique.
• p != q
• p and q will exist in the tree.
"""
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def lowestCommonAncestor(
self, root: "TreeNode", p: "TreeNode", q: "TreeNode"
) -> "TreeNode":
if not root:
return None
if root == p or root == q:
return root
l = self.lowestCommonAncestor(root.left, p, q)
r = self.lowestCommonAncestor(root.right, p, q)
if l and r:
return root
else:
return l or r