Problem Description

You have a binary tree where each node contains a specific value with the following meaning:

Leaf nodes (nodes with no children):

• Value 0 represents false
• Value 1 represents true

Non-leaf nodes (internal nodes) represent boolean operations:

• Value 2 represents OR operation
• Value 3 represents AND operation
• Value 4 represents XOR operation
• Value 5 represents NOT operation

Special note: NOT nodes have exactly one child (either left or right), while all other non-leaf nodes have exactly two children.

The tree evaluates to a boolean value following these rules:

• A leaf node evaluates to its boolean value (0 → false, 1 → true)
• A non-leaf node evaluates by first evaluating its children, then applying its boolean operation to those child values

You can perform a "flip" operation on any leaf node, which changes:

• 0 (false) to 1 (true)
• 1 (true) to 0 (false)

Given a binary tree root and a target boolean value result, find the minimum number of leaf flips needed to make the entire tree evaluate to result.

For example, if you have an AND node (value 3) with two leaf children 1 and 0, the tree evaluates to true AND false = false. To make it evaluate to true, you would need to flip the 0 leaf to 1, requiring 1 flip operation.

The problem guarantees that it's always possible to achieve the desired result through some combination of flips.

Flowchart Walkthrough

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

Is it a graph?

• Yes: While this problem involves a binary tree, a tree is a special type of graph (specifically, a connected acyclic graph). Binary trees are graphs with nodes connected by edges, where each node has at most two children.

Is it a tree?

• Yes: The problem explicitly states we're working with a binary tree. We have a root node, and each node can have up to two children (left and right). The tree structure is used to represent boolean expressions with operators at internal nodes and boolean values at leaf nodes.

DFS

• Yes: According to the flowchart, when we have a tree structure, DFS (Depth-First Search) is the recommended approach.

Conclusion: The flowchart suggests using DFS for this problem. This makes perfect sense because:

1. We need to traverse the entire tree to evaluate boolean expressions
2. We must compute values bottom-up (from leaves to root) to determine how each subtree evaluates
3. At each node, we need to consider the minimum flips required to achieve both true and false results
4. DFS allows us to recursively process each subtree and combine results at parent nodes based on the boolean operation

The DFS approach naturally handles the recursive nature of evaluating boolean expressions in a tree, where each non-leaf node's value depends on its children's evaluated values. We can use DFS to compute, for each node, the minimum flips needed to make that subtree evaluate to either true or false, then select the appropriate value based on the desired result.

Intuition

The key insight is that at any node in the tree, we need to know two pieces of information: how many flips are needed to make this subtree evaluate to false, and how many flips are needed to make it evaluate to true. Why both? Because the parent node might need either value depending on its operation and what result it's trying to achieve.

Think about it this way: if you're at an AND node and want the result to be true, you need both children to be true. But if you're at an OR node and want true, you only need one child to be true. So each child subtree should tell us: "It costs X flips to make me false and Y flips to make me true."

For leaf nodes, this is straightforward:

• If the leaf is 0 (false), it costs 0 flips to be false and 1 flip to be true
• If the leaf is 1 (true), it costs 1 flip to be false and 0 flips to be true

For non-leaf nodes, we combine the costs from children based on the boolean operation:

• OR node (2): To get false, both children must be false. To get true, at least one child must be true (so we pick the cheapest option).

• AND node (3): To get false, at least one child must be false (pick the cheapest). To get true, both children must be true.

• XOR node (4): To get false, children must have the same value (both true or both false). To get true, children must differ.

• NOT node (5): Simply inverts the child - to get false, we need the child to be true, and vice versa.

By computing both possibilities (cost to achieve false and cost to achieve true) at each node bottom-up, we can propagate this information to the root. At the root, we simply pick the cost corresponding to our desired result.

This dynamic programming approach on the tree ensures we find the minimum number of flips by always choosing the cheapest option at each decision point while traversing from leaves to root.

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

Solution Approach

The implementation uses tree dynamic programming with DFS to compute the minimum flips needed. We define a recursive function dfs(root) that returns a tuple (cost_to_false, cost_to_true) representing the minimum flips needed to make the subtree rooted at root evaluate to false and true respectively.

Base Case:

• If the node is None, return (inf, inf) since an empty tree can't be evaluated.

Leaf Nodes (values 0 or 1):

• For value 0 (false): return (0, 1) - costs 0 flips to remain false, 1 flip to become true
• For value 1 (true): return (1, 0) - costs 1 flip to become false, 0 flips to remain true
• This is implemented as return (x, x ^ 1) where x is the node value

Non-Leaf Nodes: First, recursively compute costs for left and right children:

l, r = dfs(root.left), dfs(root.right)

Then handle each boolean operation:

1. OR Operation (value = 2):

   • To get false: Both children must be false → l[0] + r[0]
   • To get true: At least one child must be true → min(l[0] + r[1], l[1] + r[0], l[1] + r[1])

2. AND Operation (value = 3):

   • To get false: At least one child must be false → min(l[0] + r[0], l[0] + r[1], l[1] + r[0])
   • To get true: Both children must be true → l[1] + r[1]

3. XOR Operation (value = 4):

   • To get false: Children must have same value → min(l[0] + r[0], l[1] + r[1])
   • To get true: Children must have different values → min(l[0] + r[1], l[1] + r[0])

4. NOT Operation (value = 5):

   • To get false: Child must be true → min(l[1], r[1]) (only one child exists)
   • To get true: Child must be false → min(l[0], r[0])

Final Result: After computing the costs for the root, we return dfs(root)[int(result)]:

• If result is False, we need dfs(root)[0]
• If result is True, we need dfs(root)[1]

The time complexity is O(n) where n is the number of nodes, as we visit each node once. The space complexity is O(h) where h is the height of the tree, due to the recursion stack.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Consider this binary tree where we want the result to be True:

       AND (3)
       /      \
    OR (2)    true (1)
    /    \
false(0) true(1)

We'll compute the costs bottom-up using DFS.

Step 1: Process the leftmost leaf false (0)

• Cost to make it false: 0 flips (already false)
• Cost to make it true: 1 flip
• Returns: (0, 1)

Step 2: Process the middle leaf true (1)

• Cost to make it false: 1 flip
• Cost to make it true: 0 flips (already true)
• Returns: (1, 0)

Step 3: Process the OR node (2)

• Left child costs: (0, 1)
• Right child costs: (1, 0)
• For OR to be false: both children must be false → 0 + 1 = 1 flip
• For OR to be true: at least one child must be true → min(0+0, 1+1, 1+0) = 0 flips
• Returns: (1, 0)

Step 4: Process the rightmost leaf true (1)

• Cost to make it false: 1 flip
• Cost to make it true: 0 flips
• Returns: (1, 0)

Step 5: Process the root AND node (3)

• Left child (OR) costs: (1, 0)
• Right child costs: (1, 0)
• For AND to be false: at least one child must be false → min(1+1, 1+0, 0+1) = 1 flip
• For AND to be true: both children must be true → 0 + 0 = 0 flips
• Returns: (1, 0)

Final Result: Since we want the tree to evaluate to True, we look at index 1 of the root's result: 0 flips needed.

This makes sense: the OR node already evaluates to true (false OR true = true), and the rightmost leaf is already true, so true AND true = true with no flips needed!

If we wanted the result to be False instead, we'd need 1 flip (either flip the middle leaf from true to false, making the OR evaluate to false, or flip the rightmost leaf from true to false).

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm performs a depth-first search (DFS) traversal of the binary tree. Each node is visited exactly once during the traversal. At each node, the function performs constant time operations:

• For leaf nodes (values 0 or 1), it returns values in O(1) time
• For internal nodes (values 2, 3, 4, or 5), it performs constant time calculations using the results from child nodes

Since we visit all n nodes exactly once and perform O(1) work at each node, the total time complexity is O(n).

Space Complexity: O(n)

The space complexity comes from the recursive call stack used during the DFS traversal. In the worst case, when the tree is completely unbalanced (essentially a linked list), the recursion depth can be O(n). In the best case of a balanced tree, the recursion depth would be O(log n). However, considering the worst-case scenario, the space complexity is O(n).

Additionally, each recursive call stores a constant amount of data (the return values and local variables), which doesn't change the overall space complexity of O(n).

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

Common Pitfalls

1. Incorrect Handling of NOT Operation's Single Child

Pitfall: A common mistake is assuming that NOT nodes always have their child in a specific position (either left or right). Developers often write code like:

# WRONG: Assumes child is always on the left
if node_value == 5:
    cost_to_false = left_costs[1]
    cost_to_true = left_costs[0]

Problem: If the NOT node's only child is on the right side, left_costs would be (inf, inf), leading to incorrect results.

Solution: Use min() to handle both possibilities:

# CORRECT: Handles child on either side
if node_value == 5:
    cost_to_false = min(left_costs[1], right_costs[1])
    cost_to_true = min(left_costs[0], right_costs[0])

2. Forgetting to Consider All Combinations for OR/AND Operations

Pitfall: When calculating the minimum cost for OR to evaluate to true (or AND to evaluate to false), developers might forget some valid combinations:

# WRONG: Missing the case where both children are true
if node_value == 2:  # OR
    cost_to_true = min(
        left_costs[0] + right_costs[1],
        left_costs[1] + right_costs[0]
    )  # Missing: left_costs[1] + right_costs[1]

Problem: This misses valid scenarios and may return a suboptimal answer.

Solution: Include all valid combinations:

# CORRECT: Includes all cases where OR evaluates to true
cost_to_true = min(
    left_costs[0] + right_costs[1],  # F OR T = T
    left_costs[1] + right_costs[0],  # T OR F = T
    left_costs[1] + right_costs[1]   # T OR T = T
)

3. Confusing Leaf Node Return Values

Pitfall: Incorrectly calculating the flip costs for leaf nodes:

# WRONG: Reversed logic
if node_value in (0, 1):
    return 1 - node_value, node_value  # Incorrect!

Problem: This reverses the costs - a node with value 0 would return (1, 0) instead of (0, 1).

Solution: Remember that the tuple represents (cost_to_make_false, cost_to_make_true):

# CORRECT: For a leaf with value x
# - Cost to make it false: x (flip if it's 1, don't flip if it's 0)
# - Cost to make it true: x^1 (flip if it's 0, don't flip if it's 1)
return node_value, node_value ^ 1

4. Not Handling Empty/None Nodes Properly

Pitfall: Returning incorrect values for None nodes or not handling them at all:

# WRONG: Returns 0 costs for None nodes
if node is None:
    return 0, 0

Problem: This would incorrectly allow operations on non-existent children, producing wrong results.

Solution: Return infinity to indicate impossibility:

# CORRECT: None nodes can't be evaluated
if node is None:
    return inf, inf

This ensures that when min() is used with NOT operations, only the existing child's cost is considered.