Problem Description

You have a binary tree with n nodes labeled from 1 to n. The tree structure follows a specific pattern:

• Node 1 is the root
• For any node with label v, its parent is the node with label floor(v / 2)
• This means node 2 and node 3 are children of node 1, nodes 4 and 5 are children of node 2, and so on

Initially, all nodes have a value of 0.

You're given an array queries where each queries[i] represents a node label. For each query, you need to flip the values in the entire subtree rooted at that node (including the node itself). Flipping means changing 0 to 1 and 1 to 0.

After processing all queries in order, you need to return the total count of nodes that have value 1.

Example of tree structure when n = 7:

• Node 1 is the root
• Node 3 has parent floor(3/2) = 1
• Node 7 has parent floor(7/2) = 3

The tree forms a complete binary tree pattern where:

• Node i has left child 2*i and right child 2*i + 1 (if they exist within range 1 to n)

Key insight from the solution: If a node appears an even number of times in the queries array, the flips cancel out (flipping twice returns to original state). Only nodes that appear an odd number of times in queries need their subtrees flipped. The solution uses Counter to count query frequencies and only processes nodes with odd counts, using DFS to flip entire subtrees efficiently.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The problem involves a tree structure with nodes and edges. Each node has a parent-child relationship defined by parent = floor(node/2), forming a binary tree which is a special type of graph.

Is it a tree?

• Yes: The problem explicitly states we have an undirected connected tree with n nodes and n-1 edges. The structure follows a binary tree pattern where node 1 is the root and each node's children are 2*node and 2*node + 1.

DFS

• We arrive at DFS as the solution approach.

Conclusion: The flowchart correctly leads us to use DFS (Depth-First Search) for this problem. This makes perfect sense because:

1. We need to traverse entire subtrees for each query operation
2. When we flip a node's value, we must recursively flip all values in its subtree
3. DFS is the natural choice for exploring all nodes in a subtree before moving to other parts of the tree
4. The solution implements DFS through the recursive dfs(i) function that:
   • Flips the current node's value with tree[i] ^= 1
   • Recursively calls itself on the left child dfs(i << 1)
   • Recursively calls itself on the right child dfs(i << 1 | 1)

The DFS pattern perfectly fits this problem as we need to perform operations on entire subtrees, which requires visiting all descendants of a given node - a classic use case for depth-first traversal.

Intuition

The key insight starts with recognizing that flipping values in a subtree is a toggle operation - doing it twice returns to the original state. If we flip a subtree an even number of times, it's as if we never flipped it at all. Only odd numbers of flips actually change the final state.

This leads to our first optimization: instead of processing queries sequentially, we can count how many times each node appears in the queries array. If node i appears 3 times, that's equivalent to flipping it once. If it appears 4 times, we don't need to flip it at all.

The second insight comes from the tree structure. The problem defines a complete binary tree where:

• Node i has left child 2*i (which is i << 1 in binary)
• Node i has right child 2*i + 1 (which is i << 1 | 1 in binary)

This binary relationship makes traversing the tree extremely efficient using bit operations.

When we need to flip a subtree, DFS is the natural choice because:

1. We start at the query node and need to visit every descendant
2. The recursive nature of DFS mirrors the recursive structure of the tree
3. We can flip the current node, then recursively flip left and right subtrees

The final piece is realizing we can use an array to track node values since nodes are labeled from 1 to n. This gives us O(1) access to any node's value.

Putting it all together: Count query frequencies → Process only nodes with odd counts → Use DFS to flip each subtree → Count the final number of 1s. The XOR operation (^= 1) elegantly handles the flip since 0 ^ 1 = 1 and 1 ^ 1 = 0.

Learn more about Tree, Depth-First Search, Breadth-First Search and Binary Tree patterns.

Solution Approach

The implementation follows the optimization strategy mentioned in the reference: only process nodes that appear an odd number of times in the queries.

Step 1: Count Query Frequencies

cnt = Counter(queries)

We use Python's Counter to count how many times each node appears in the queries array. This allows us to identify which nodes need to be processed.

Step 2: Initialize the Tree

tree = [0] * (n + 1)

Create an array of size n+1 to represent node values. Index i corresponds to node i (we ignore index 0 for simplicity). Initially, all values are 0.

Step 3: Define the DFS Function

def dfs(i):
    if i > n:
        return
    [tree](/problems/tree_intro)[i] ^= 1
    dfs(i << 1)
    dfs(i << 1 | 1)

The DFS function recursively flips values in a subtree:

• Base case: If node index exceeds n, we've gone beyond the tree bounds
• Flip current node: tree[i] ^= 1 toggles between 0 and 1
• Recursively process left child: i << 1 (equivalent to 2*i)
• Recursively process right child: i << 1 | 1 (equivalent to 2*i + 1)

Step 4: Process Nodes with Odd Query Counts

for i, v in cnt.items():
    if v & 1:
        dfs(i)

Iterate through the frequency counter. For each node i that appears v times:

• Check if v is odd using bitwise AND: v & 1 returns 1 for odd numbers, 0 for even
• If odd, call dfs(i) to flip the entire subtree rooted at node i

Step 5: Count Final Result

return sum([tree](/problems/tree_intro))

Sum all values in the tree array to get the total number of nodes with value 1.

Time Complexity: O(n × q) in worst case where q is the number of unique nodes in queries, as each DFS traversal visits all nodes in a subtree.

Space Complexity: O(n) for the tree array and O(log n) for the recursive call stack depth.

Example Walkthrough

Let's walk through a small example with n = 7 and queries = [1, 3, 3].

Initial Setup: The tree structure looks like this:

        1(0)
       /    \
     2(0)    3(0)
    /  \    /  \
   4(0) 5(0) 6(0) 7(0)

Numbers in parentheses show node values (all initially 0).

Step 1: Count Query Frequencies

cnt = Counter([1, 3, 3])  # Result: {1: 1, 3: 2}

• Node 1 appears 1 time (odd)
• Node 3 appears 2 times (even)

Step 2: Process Only Odd-Count Nodes Since node 3 appears an even number of times, its flips cancel out. We only need to process node 1.

Step 3: DFS from Node 1 Starting DFS at node 1:

dfs(1):
  - Flip node 1: 0 → 1
  - Call dfs(2) for left child
  - Call dfs(3) for right child

dfs(2):
  - Flip node 2: 0 → 1
  - Call dfs(4) for left child
  - Call dfs(5) for right child

dfs(4):
  - Flip node 4: 0 → 1
  - Call dfs(8) → returns (8 > 7)
  - Call dfs(9) → returns (9 > 7)

dfs(5):
  - Flip node 5: 0 → 1
  - Call dfs(10) → returns (10 > 7)
  - Call dfs(11) → returns (11 > 7)

dfs(3):
  - Flip node 3: 0 → 1
  - Call dfs(6) for left child
  - Call dfs(7) for right child

dfs(6):
  - Flip node 6: 0 → 1
  - Call dfs(12) → returns (12 > 7)
  - Call dfs(13) → returns (13 > 7)

dfs(7):
  - Flip node 7: 0 → 1
  - Call dfs(14) → returns (14 > 7)
  - Call dfs(15) → returns (15 > 7)

Final Tree State:

        1(1)
       /    \
     2(1)    3(1)
    /  \    /  \
   4(1) 5(1) 6(1) 7(1)

Step 4: Count Result

sum(tree) = 7  # All nodes have value 1

The key optimization: We avoided processing node 3's subtree twice (which would have flipped it back to 0) by recognizing it appears an even number of times in the queries.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n × log n)

The algorithm iterates through unique queries in the Counter. For each query that appears an odd number of times, it calls the dfs function starting from that node. In the worst case, the dfs function traverses from a node down to all its descendants in the complete binary tree.

For a complete binary tree with n nodes:

• The maximum depth is O(log n)
• From any node i, the number of nodes in its subtree (including itself) is at most n/i
• When we sum up all possible DFS traversals, considering that each node can be visited multiple times from different ancestors, the total operations across all queries is bounded by O(n × log n)

This is because each node at depth d can be reached from at most d ancestors (nodes on the path from root to this node), and with maximum depth being O(log n), the total work done is O(n × log n).

Space Complexity: O(n)

The space complexity consists of:

• The tree array of size n + 1: O(n)
• The Counter object storing unique queries: O(min(len(queries), n)) which is bounded by O(n)
• The recursive call stack for dfs: O(log n) for the maximum depth of recursion

The dominant term is O(n) from the tree array, giving us an overall space complexity of O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Not Handling the Tree Index Bounds Correctly

A frequent mistake is forgetting that the tree is 1-indexed while also having a maximum of n nodes. This can lead to two types of errors:

Pitfall Example:

# Wrong: Using 0-indexed array
tree = [0] * n  # Missing one element!

# Wrong: Incorrect boundary check
def dfs(i):
    if i >= n:  # Should be i > n
        return
    tree[i] ^= 1
    # ...

Solution:

# Correct: Account for 1-indexing
tree = [0] * (n + 1)  # Extra element for 1-indexing

def dfs(i):
    if i > n:  # Correct boundary check
        return
    tree[i] ^= 1
    # ...

2. Processing All Queries Instead of Optimizing

The naive approach of processing each query individually leads to unnecessary work when the same node appears multiple times:

Pitfall Example:

# Inefficient: Processing each query separately
for query in queries:
    dfs(query)
return sum(tree)

This approach performs redundant flips when a node appears an even number of times (the flips cancel out).

Solution:

# Efficient: Only process nodes with odd occurrence count
cnt = Counter(queries)
for node, frequency in cnt.items():
    if frequency % 2 == 1:  # Only odd counts matter
        dfs(node)

3. Using Incorrect Child Node Calculation

In a complete binary tree, the child indices must be calculated precisely:

Pitfall Example:

def dfs(i):
    if i > n:
        return
    tree[i] ^= 1
    dfs(i * 2 - 1)  # Wrong: This isn't the left child
    dfs(i * 2)      # Wrong: This would be left child, not right

Solution:

def dfs(i):
    if i > n:
        return
    tree[i] ^= 1
    dfs(i * 2)      # Correct: Left child
    dfs(i * 2 + 1)  # Correct: Right child
    # Or using bitwise operations:
    # dfs(i << 1)     # Left child
    # dfs(i << 1 | 1) # Right child

4. Stack Overflow for Large Trees

For very large values of n, the recursive DFS might cause stack overflow:

Solution using iterative approach:

def toggle_subtree_iterative(root):
    stack = [root]
    while stack:
        node = stack.pop()
        if node > n:
            continue
        tree[node] ^= 1
        stack.append(node * 2)
        stack.append(node * 2 + 1)

5. Misunderstanding the Flip Operation

Some might incorrectly implement the flip as setting to 1 rather than toggling:

Pitfall Example:

def dfs(i):
    if i > n:
        return
    tree[i] = 1  # Wrong: This sets to 1, doesn't toggle!
    # ...

Solution:

def dfs(i):
    if i > n:
        return
    tree[i] ^= 1  # Correct: XOR toggles between 0 and 1
    # Or alternatively:
    # tree[i] = 1 - tree[i]