Problem Description

This problem asks you to create a visual representation of a binary tree as a 2D string matrix following specific formatting rules.

Given a binary tree, you need to construct a matrix where:

1. Matrix dimensions:

   • The number of rows m equals height + 1, where height is the height of the tree
   • The number of columns n equals 2^(height+1) - 1

2. Node placement rules:

   • The root node goes in the middle of the first row at position res[0][(n-1)/2]
   • For any node placed at position res[r][c]:
     • Its left child goes at res[r+1][c - 2^(height-r-1)]
     • Its right child goes at res[r+1][c + 2^(height-r-1)]
   • Node values should be converted to strings when placed in the matrix
   • Empty positions should contain empty strings ""

The solution uses two main steps:

1. Calculate tree height: A recursive function finds the maximum depth of the tree. The height is defined as the number of edges in the longest path from root to leaf.

2. Fill the matrix: Using DFS traversal starting from the root, each node is placed at its calculated position. The formula 2^(height-r-1) determines the horizontal offset for child nodes, which decreases as you go deeper in the tree (larger r values).

The key insight is that the offset for placing children nodes depends on both the tree's total height and the current row number. This ensures proper spacing that creates a balanced visual representation where each level's nodes are evenly distributed across the available columns.

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 not explicitly a graph problem, a binary tree is a special type of graph (specifically, a connected acyclic graph).

Is it a tree?

• Yes: The problem explicitly deals with a binary tree structure where we need to traverse and process each node.

DFS

• Yes: This is our conclusion. The problem requires visiting each node in the tree to place it in the correct position in the output matrix.

Conclusion: The flowchart suggests using DFS (Depth-First Search) for this problem.

This makes sense because:

1. We need to traverse the entire tree structure to place each node in the matrix
2. DFS allows us to recursively process each node along with its position information (row and column)
3. The parent-child relationship in the tree naturally maps to the recursive nature of DFS, where we process a node and then recursively process its children with updated position parameters
4. The formula for calculating child positions (c - 2^(height-r-1) for left, c + 2^(height-r-1) for right) requires maintaining the current depth/row, which DFS handles elegantly through its recursive call stack

Intuition

The key insight is to think about how we would draw a binary tree on paper. We naturally place the root at the top center, then position children nodes below their parents with appropriate spacing to avoid overlaps.

To formalize this intuitive drawing process, we need to determine:

1. How large should our canvas (matrix) be?
2. Where exactly should each node be placed?

For the matrix size, we observe that:

• Each level of the tree needs its own row, so we need height + 1 rows
• The width must accommodate the maximum possible spread of nodes. In a complete binary tree of height h, the bottom level can have up to 2^h nodes, and we need spaces between them. This leads to 2^(h+1) - 1 columns.

For node positioning, imagine dividing the available horizontal space recursively:

• The root takes the middle position
• Each child gets half of its parent's "territory"
• The offset from parent to child decreases by half at each level

This halving pattern gives us the formula 2^(height-r-1) for the offset at row r. At the top (row 0), children are placed far apart with offset 2^height. As we go deeper, the offset decreases: 2^(height-1), 2^(height-2), and so on, until leaves have an offset of 2^0 = 1.

The DFS approach naturally fits because:

• We process each node once (place its value in the matrix)
• We need to pass down contextual information (current position) to children
• The recursive structure mirrors the tree structure itself

By calculating the tree height first, we can determine both the matrix dimensions and the correct offsets for each level, then use DFS to systematically fill in each node's value at its calculated position.

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

Solution Approach

The implementation consists of two main phases: calculating the tree height and filling the matrix using DFS.

Phase 1: Calculate Tree Height

The height function recursively computes the tree's height:

• Base case: If the node is None, return -1 (this ensures a single node has height 0)
• Recursive case: Return 1 + max(height(left), height(right))

This gives us the maximum depth from root to any leaf, which we need for determining matrix dimensions and calculating offsets.

Phase 2: Matrix Construction and DFS Traversal

Once we have the height h, we:

1. Calculate matrix dimensions:
   • Rows: m = h + 1
   • Columns: n = 2^(h+1) - 1
2. Initialize an m × n matrix filled with empty strings ""
3. Start DFS from the root at position (0, (n-1)//2)

The dfs function places nodes in the matrix:

def dfs(root, r, c):
    if root is None:
        return
    ans[r][c] = str(root.val)  # Place current node's value
    dfs(root.left, r + 1, c - 2^(h - r - 1))   # Place left child
    dfs(root.right, r + 1, c + 2^(h - r - 1))  # Place right child

Key Implementation Details:

• Position Calculation: The offset 2^(h - r - 1) determines how far left or right to place children from their parent

  • At row 0: offset = 2^h (maximum spacing)
  • At row 1: offset = 2^(h-1) (half the previous)
  • This pattern continues, halving at each level

• String Conversion: Node values are converted to strings using str(root.val) before placement

• Matrix Initialization: Using list comprehension [[""] * n for _ in range(m)] creates the empty matrix efficiently

The algorithm visits each node exactly once, making it O(n) in time complexity where n is the number of nodes. The space complexity is O(m × n) for the output matrix, which is O(2^h × h) in terms of tree height.

Example Walkthrough

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

Consider this binary tree:

    1
   / \
  2   3
   \
    4

Step 1: Calculate Tree Height

Starting from the root (1):

• height(1) = 1 + max(height(2), height(3))
  • height(2) = 1 + max(height(None), height(4))
    • height(4) = 1 + max(height(None), height(None)) = 1 + max(-1, -1) = 0
    • So height(2) = 1 + max(-1, 0) = 1
  • height(3) = 1 + max(height(None), height(None)) = 1 + max(-1, -1) = 0
  • So height(1) = 1 + max(1, 0) = 2

The tree height is 2.

Step 2: Calculate Matrix Dimensions

• Rows: m = height + 1 = 2 + 1 = 3 rows
• Columns: n = 2^(height+1) - 1 = 2^3 - 1 = 7 columns

Initialize a 3×7 matrix filled with empty strings:

[["", "", "", "", "", "", ""],
 ["", "", "", "", "", "", ""],
 ["", "", "", "", "", "", ""]]

Step 3: DFS Traversal to Fill Matrix

Start DFS from root at position (0, (7-1)/2) = (0, 3):

1. Place root (1) at row 0, col 3:

   • ans[0][3] = "1"
   • Calculate offset: 2^(2-0-1) = 2^1 = 2
   • Process left child (2) at position (1, 3-2) = (1, 1)
   • Process right child (3) at position (1, 3+2) = (1, 5)

2. Place node 2 at row 1, col 1:

   • ans[1][1] = "2"
   • Calculate offset: 2^(2-1-1) = 2^0 = 1
   • Left child is None (skip)
   • Process right child (4) at position (2, 1+1) = (2, 2)

3. Place node 3 at row 1, col 5:

   • ans[1][5] = "3"
   • Both children are None (skip)

4. Place node 4 at row 2, col 2:

   • ans[2][2] = "4"
   • Both children are None (skip)

Final Result:

[["", "", "", "1", "", "", ""],
 ["", "2", "", "", "", "3", ""],
 ["", "", "4", "", "", "", ""]]

The matrix shows:

• Root "1" centered in row 0
• "2" and "3" in row 1, positioned with offset 2 from their parent
• "4" in row 2, positioned with offset 1 from its parent "2"

Each level's offset halves as we go deeper: starting at 2^1=2 for level 1, then 2^0=1 for level 2, ensuring proper spacing throughout the tree visualization.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n * 2^h) where n is the number of nodes in the tree and h is the height of the tree.

• The height function visits every node once to calculate the height, taking O(n) time.
• The dfs function visits each node in the tree exactly once, performing O(1) work at each node (assigning a value to the matrix), taking O(n) time.
• Creating the answer matrix with dimensions (h+1) × (2^(h+1) - 1) and initializing all cells with empty strings takes O(h * 2^h) time.
• Since in the worst case (a complete binary tree), n = 2^(h+1) - 1, the matrix initialization dominates, giving us O(n * 2^h) time complexity.
• In a balanced tree, h = O(log n), so this becomes O(n^2).
• In a skewed tree, h = O(n), so this becomes O(n * 2^n).

Space Complexity: O(h * 2^h) or equivalently O(m * n) where m and n are the dimensions of the output matrix.

• The output matrix ans has dimensions (h+1) × (2^(h+1) - 1), requiring O(h * 2^h) space.
• The recursion stack depth for both height and dfs functions is at most O(h).
• Since the output matrix space dominates, the overall space complexity is O(h * 2^h).
• In terms of the number of nodes, for a balanced tree where h = O(log n), this becomes O(n log n).
• For a skewed tree where h = O(n), this becomes O(n * 2^n).

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

Common Pitfalls

1. Incorrect Height Calculation

A common mistake is defining the height function incorrectly, particularly the base case. Some implementations might return 0 for a None node, which would make a single-node tree have height 1 instead of 0.

Incorrect:

def get_tree_height(node):
    if node is None:
        return 0  # Wrong! This makes a leaf node have height 1
    return 1 + max(get_tree_height(node.left), get_tree_height(node.right))

Correct:

def get_tree_height(node):
    if node is None:
        return -1  # Correct! This makes a leaf node have height 0
    return 1 + max(get_tree_height(node.left), get_tree_height(node.right))

2. Off-by-One Error in Offset Calculation

The offset formula 2^(height - row - 1) is crucial. A common error is forgetting the -1 or miscalculating the exponent, leading to incorrect child placement.

Incorrect:

offset = 2 ** (tree_height - row)  # Missing the -1
# or
offset = 2 ** (row)  # Completely wrong formula

Correct:

offset = 2 ** (tree_height - row - 1)

3. Integer Division vs Floor Division

When calculating the center column for the root, using regular division instead of floor division can cause issues in Python 3.

Incorrect:

center_col = (num_cols - 1) / 2  # Returns float in Python 3
result[0][center_col] = str(root.val)  # TypeError: list indices must be integers

Correct:

center_col = (num_cols - 1) // 2  # Integer division

4. Matrix Initialization with Reference Issues

Creating the matrix incorrectly can lead to all rows referencing the same list object.

Incorrect:

result = [[""] * num_cols] * num_rows  # All rows reference the same list!
# Modifying result[0][0] would affect all rows

Correct:

result = [[""] * num_cols for _ in range(num_rows)]  # Each row is a separate list

5. Forgetting to Convert Node Values to Strings

The problem requires string values in the matrix, but it's easy to forget the conversion.

Incorrect:

result[row][col] = node.val  # Mixing integers and strings in the result

Correct:

result[row][col] = str(node.val)

6. Not Handling Edge Cases

Failing to consider special cases like empty trees or single nodes.

Solution: Always test with:

• Empty tree (though most problems guarantee at least one node)
• Single node tree (height = 0)
• Skewed trees (all nodes on one side)
• Complete binary trees