Given the root of a binary tree, construct a 0-indexed m x n string matrix res that represents a formatted layout of the tree. The formatted layout matrix should be constructed using the following rules:
height and the number of rows m should be equal to height + 1.n should be equal to 2height+1 - 1.res[0][(n-1)/2]).res[r][c], place its left child at res[r+1][c-2height-r-1] and its right child at res[r+1][c+2height-r-1]."".Return the constructed matrix res.
Example 1:
Input: root = [1,2] Output: [["","1",""], ["2","",""]]
Example 2:
Input: root = [1,2,3,null,4] Output: [["","","","1","","",""], ["","2","","","","3",""], ["","","4","","","",""]]
Constraints:
[1, 210].-99 <= Node.val <= 99[1, 10].When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The basic idea is to represent the tree as a grid of text and fill in the nodes level by level. We will try every possible position for each node in the tree to find the best fit. This is a very simple and inefficient way to print the tree.
Here's how the algorithm would work step-by-step:
class Node:
def __init__(self, value, left=None, right=None):
self.value = value
self.left = left
self.right = right
def print_binary_tree_brute_force(root):
if not root:
return
# Determine tree height for grid size
tree_height = get_tree_height(root)
grid_width = 2**(tree_height - 1) * 2 - 1
grid_height = tree_height
# Initialize the grid with empty strings
grid = [['' for _ in range(grid_width)] for _ in range(grid_height)]
all_arrangements = []
def explore_arrangements(node, row, col, current_arrangement):
if not node:
return
# Try placing the node at the current position
current_arrangement[row][col] = str(node.value)
#Explore subtrees
left_col = col - 2**(tree_height - row - 2) if node.left else None
right_col = col + 2**(tree_height - row - 2) if node.right else None
if node.left:
explore_arrangements(node.left, row + 1, left_col, current_arrangement)
if node.right:
explore_arrangements(node.right, row + 1, right_col, current_arrangement)
def is_valid(arrangement):
# Check for overlaps in arrangement
for row in arrangement:
last_col = -1
for col_index, cell in enumerate(row):
if cell:
if last_col != -1 and col_index - last_col < 2:
return False
last_col = col_index
return True
def generate_arrangements(node, row):
if not node:
return []
valid_arrangements = []
for col in range(grid_width):
new_grid = [['' for _ in range(grid_width)] for _ in range(grid_height)]
explore_arrangements(node, row, col, new_grid)
if is_valid(new_grid):
valid_arrangements.append(new_grid)
return valid_arrangements
# Generate all possible valid arrangements
all_arrangements = generate_arrangements(root, 0)
# Pick the 'best' arrangement
best_arrangement = None
if all_arrangements:
best_arrangement = all_arrangements[0]
# Print the 'best' arrangement
if best_arrangement:
for row in best_arrangement:
print(''.join(['{:3}'.format(cell) for cell in row]))
else:
print("No valid arrangement found.")
def get_tree_height(node):
if not node:
return 0
return 1 + max(get_tree_height(node.left), get_tree_height(node.right))To print a binary tree in a structured way, we need to figure out each node's position in a grid. We'll use a smart approach that figures out the tree's dimensions and then fills the grid level by level, placing each node where it belongs.
Here's how the algorithm would work step-by-step:
def print_binary_tree(root):
def get_tree_height(node):
if not node:
return 0
return 1 + max(get_tree_height(node.left), get_tree_height(node.right))
def get_tree_width(height):
return 2**height - 1
height_of_tree = get_tree_height(root)
width_of_tree = get_tree_width(height_of_tree)
tree_grid = [[''] * width_of_tree for _ in range(height_of_tree)]
def populate_grid(node, row, left, right):
if not node:
return
mid = (left + right) // 2
tree_grid[row][mid] = str(node.val)
# Recursively populate the left subtree.
populate_grid(node.left, row + 1, left, mid - 1)
# Recursively populate the right subtree.
populate_grid(node.right, row + 1, mid + 1, right)
# Populate the grid starting from the root.
populate_grid(root, 0, 0, width_of_tree - 1)
for row in tree_grid:
print(''.join(row))
class TreeNode:
def __init__(self, val=0, left=None, right=None):
self.val = val
self.left = left
self.right = right| Case | How to Handle |
|---|---|
| Null or empty tree | Return an empty list immediately, as there is nothing to print. |
| Tree with only a root node | Print the root node's value in a list containing just that value. |
| Completely unbalanced tree (left-skewed) | The algorithm should still correctly print the tree, although the visual layout might be very wide if a level-order traversal is used. |
| Completely unbalanced tree (right-skewed) | Similar to the left-skewed case, the algorithm should handle it correctly but with potential visual layout issues. |
| Tree with duplicate values | The printing algorithm should handle duplicate values without issues, printing each instance of the value as it appears in the tree's structure. |
| Tree with negative values | The algorithm should be able to handle nodes with negative values. |
| Very large tree (potential stack overflow with recursion) | If using a recursive approach, consider the stack depth and potentially switch to an iterative approach like level-order traversal to prevent stack overflow. |
| Integer overflow in node values (if applicable to the problem context) | Be mindful of potential integer overflows when performing calculations with node values and use appropriate data types or overflow checks. |