# 5. Trees and Tries

## A. General Introduction

### I. Key Methods

1. Identify the base case - what to do when it reaches the leaf or root node?
2. Recursive calls to each child node
3. Identify what needs to be calculated with the information from the child nodes
4. Return the calculated result

### II. Binary Search Tree (BST) Properties

1. Inorder traversal gives the sorted order of the BST
2. All values in left subtree are smaller than the current node
3. All values in right subtree are larger than the current node

## B. Leetcode problems

### 226. Invert Binary Tree

Given the root of a binary tree, invert the tree, and return its root.

##### Key Ideas
• Invert the tree by swapping the left and right children of each node.
Code


# 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 root is None or root.left == None and root.right == None:
return root

root.left, root.right = root.right, root.left
self.invertTree(root.left)
self.invertTree(root.right)

return root



### 543. Diameter of Binary Tree

Given the root of a binary tree, return the length of the diameter of the tree.

The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.

The length of a path between two nodes is represented by the number of edges between them.

#### Key Ideas

• @BaseCase The diameter of an empty node is 0.
• @RecursiveCase The diameter of a node is the maximum of the diameter of its left and right subtrees.
• Return the max height of the subtrees plus 1 (accounts for the edge that connects it to parent node).
code

# 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 diameterOfBinaryTree(self, root: Optional[TreeNode]) -> int:

result = 0 # global state that is set at each level

def traverse(root:TreeNode):

if not root:
return 0

rh = traverse(root.right)
lh = traverse(root.left)

nonlocal result
result = max(result, rh+lh)

return max(lh, rh) + 1

traverse(root)

return result



### 104. 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:

Input: root = [3,9,20,null,null,15,7]
Output: 3

#### Key Ideas

• @BaseCase The depth of an empty node is 0.
• @RecursiveCase Check the max depth of right and left subtrees.
• Return the max of the two plus 1 (To include the height of the current node).
code


class Solution:
def maxDepth(self, root: Optional[TreeNode]) -> int:

if not root:
return 0

depth = 0
if root.right:
rd = self.maxDepth(root.right)
depth = max(depth, rd)

if root.left:
ld = self.maxDepth(root.left)
depth = max(depth, ld)

return depth + 1



### 110. Balanced Binary Tree

Given a binary tree, determine if it is height-balanced.

For this problem, a height-balanced binary tree is defined as:

a binary tree in which the left and right subtrees of every node differ in height by no more than 1.

Example:

Input: root = [3,9,20,null,null,15,7]
Output: true

#### Key Ideas

• @BaseCase The height of an empty node is zero.
• @RecursiveCase The height of a node is the maximum of the height of its left and right subtrees.
• At every node, calculate if the difference in height between the left and right subtrees is greater than 1.
• If so, set the balanced set to false.
• Return the max height of the subtrees plus 1.
code


# 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 isBalanced(self, root: Optional[TreeNode]) -> bool:

balanced = True

def dfs(root:TreeNode) -> int:

if root is None:
return 0

lh = dfs(root.left)
rh = dfs(root.right)

if abs(lh-rh) > 1:
nonlocal balanced
balanced = False

return max(lh, rh) + 1

dfs(root)

return balanced



### 100. Same Tree

Given the roots of two binary trees p and q, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical, and the nodes have the same value.

Example:

Input: p = [1,2,3], q = [1,2,3]
Output: true

#### Key Ideas

• @BaseCase If both trees are empty, return true.
• @RecursiveCase If the trees are not empty, compare the values of the root nodes.
• If they are not equal, return false.
• Otherwise, compare the left and right subtrees.
code


class Solution:
def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:

if not p and not q:
return True

if p and q and p.val == q.val:

l = self.isSameTree(p.left, q.left)
r = self.isSameTree(p.right, q.right)

if l and r:
return True

return False



### 572. Subtree of Another Tree

Given the roots of two binary trees root and subRoot, return true if there is a subtree of root with the same structure and node values of subRoot and false otherwise.

A subtree of a binary tree tree is a tree that consists of a node in tree and all of this node’s descendants. The tree tree could also be considered as a subtree of itself.

Example:

Input: root = [3,4,5,1,2], subRoot = [4,1,2]
Output: true

#### Key Ideas

• Use isSameTree to check the equality of the trees
• @BaseCase If root is empty, return false; because there is no subtree in an empty tree.
• @BaseCase If subRoot is empty, return true; because an empty subtree is a subtree of a tree with even one node
• @RecursiveCase Check if left subtree or right subtree is the subRoot
Code

class Solution:

def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:

if not p and not q:
return True

if p and q and p.val == q.val:

l = self.isSameTree(p.left, q.left)
r = self.isSameTree(p.right, q.right)

if l and r:
return True

return False

def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:

if not root:
return False

if not subRoot:
return True

if self.isSameTree(root, subRoot):
return True

return self.isSubtree(root.left, subRoot) or self.isSubtree(root.right, subRoot)



### 102. Binary Tree Level Order Traversal

Given the root of a binary tree, return the level order traversal of its nodes' values. (i.e., from left to right, level by level).

Example:

Input: root = [3,9,20,null,null,15,7]
Output: [,[9,20],[15,7]]

#### Key Ideas

• @BaseCase If the root is empty, return an empty list.
• Used a multi-source BFS approach to traverse each level of the tree
code


class Solution:
def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:

if not root:
return []

result = []
queue = deque([root])

while queue:

layer = []
m = len(queue)

for _ in range(m):

node = queue.popleft()
layer.append(node.val)

if node.left:
queue.append(node.left)

if node.right:
queue.append(node.right)

result.append(layer)

return result



### 199. Binary Tree Right Side View

Given the root of a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.

Example:

Input: root = [1,2,3,null,5,null,4]
Output: [1, 3, 4]

#### Key Ideas

• @BaseCase If the root is empty, return an empty list.
• Used a multi-source BFS approach to traverse each level of the tree
• for each level, have a value that is overwritten each node in the level
• At the end of the loop, the value would be the last node at the level
code


class Solution:
def rightSideView(self, root: Optional[TreeNode]) -> List[int]:

if not root:
return []

result = []
queue = deque([root])

while queue:

m = len(queue)
val = None

for _ in range(m):

node = queue.popleft()
val = node.val

if node.left:
queue.append(node.left)

if node.right:
queue.append(node.right)

if val is not None:
result.append(val)

return result



### 1448. Count Good Nodes in Binary Tree

Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with a value greater than X.

Return the number of good nodes in the binary tree.

Example:

Input: root = [3,1,4,3,null,1,5]
Output: 4
Explanation: Nodes in blue are good.
Root Node (3) is always a good node.
Node 4 -> (3,4) is the maximum value in the path starting from the root.
Node 5 -> (3,4,5) is the maximum value in the path.
Node 3 -> (3,1,3) is the maximum value in the path.

#### Key Ideas:

• @MainCase Pass a maxval argument which is the maximum value in the path
• If the current node’s value is greater than the maxval, then the node is good and the maxval is updated to the current node’s value
• Traverse the right and left subtrees
Code


class Solution:
def goodNodes(self, root: TreeNode) -> int:

if not root.left and not root.right:
return 1

result = 0

def dfs(root:TreeNode, maxval:int) -> None:

if root.val >= maxval:
nonlocal result

result += 1
maxval = root.val

if root.right:
dfs(root.right, maxval)

if root.left:
dfs(root.left, maxval)

dfs(root, root.val)

return result



### 98. Validate Binary Search Tree

Given the root of a binary tree, determine if it is a valid binary search tree (BST).

A valid BST is defined as follows:

• The left subtree of a node contains only nodes with keys less than the node’s key.
• The right subtree of a node contains only nodes with keys greater than the node’s key.
• Both the left and right subtrees must also be binary search trees.

Example:

Input: root = [2,1,3]
Output: true

#### Key Ideas:

• @BaseCase If the root is empty, return True.
• Have a max and min value at each node; If the value of the node is greater than the max value, or less than the min value, then the tree is not a BST
• Check the left and right subtrees
code


class Solution:
def isValidBST(self, root: Optional[TreeNode]) -> bool:

if not root.right and not root.left:
return True

def dfs(root:TreeNode, maxval:float, minval:float):

if not root:
return True

if root.val >= maxval or root.val <= minval:
return False

r = dfs(root.right, maxval, root.val)
l = dfs(root.left, root.val, minval)

return r and l

result = dfs(root, float("inf"), float("-inf"))

return result



### 230. Kth Smallest Element in a BST

Given the root of a binary search tree, and an integer k, return the kth smallest value (1-indexed) of all the values of the nodes in the tree.

Example:

Input: root = [3,1,4,null,2], k = 1
Output: 1 \

#### Key Ideas:

• Traverse through the tree in order and return the kth value
• Return the k-1th value as it is 1-indexed
code


class Solution:
def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:

heap = []

def traverse(root: TreeNode):

if not root:
return

if root.left:
traverse(root.left)

heap.append(root.val)

if root.right:
traverse(root.right)

traverse(root)

return heap[k-1]



### 105. Construct Binary Tree from Preorder and Inorder Traversal

Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

Example:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]

#### Key Ideas:

• @BaseCase If the length of the preorder and inorder is 0, return None
• In preorder the first element is the root of the tree/subtree
• Find the index of the root in inorder and split the preorder and inorder into two subarrays
• Recursively call the function on the subarrays
code


class Solution:

def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:

if not preorder and not inorder:
return None

root = TreeNode(preorder)
rootIdx = inorder.index(preorder)

root.left = self.buildTree(preorder[1:rootIdx+1], inorder[:rootIdx])
root.right = self.buildTree(preorder[rootIdx+1:], inorder[rootIdx+1:])

return root



### 124. Binary Tree Maximum Path Sum

A path in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence at most once. Note that the path does not need to pass through the root.

The path sum of a path is the sum of the node’s values in the path.

Given the root of a binary tree, return the maximum path sum of any non-empty path.

Example:

Input: root = [-10,9,20,null,null,15,7]
Output: 42 Explanation: 5 + 20 + 7 = 42

#### Key Ideas:

• @BaseCase If the root is empty, return 0
• The idea is very similar to finding diameter of a tree - we can find the maximum path sum of any path in the tree
• Initalize the max path with root value
• Traverse through the tree and at each node, the max value can be maxvalue or <right-subtree-cost>+ root.val+<left-subtree-cost>
• Return the sum of maximum of left and right subtree max values and the root value
Code


class Solution:

def maxPathSum(self, root: Optional[TreeNode]) -> int:

maxPathVal = root.val

if not root.left and not root.right:
return root.val

def dfs(root) -> None:

if not root:
return 0

lp = max(0, dfs(root.left))
rp = max(0, dfs(root.right))

nonlocal maxPathVal
maxPathVal = max(maxPathVal, lp+rp+root.val)

return max(lp,rp) + root.val

dfs(root)

return maxPathVal



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