Leetcode 894: All Possible Full Binary Trees
Given an integer n, return all possible full binary trees with exactly n nodes. Each node of the tree must have the value 0. A full binary tree is defined as a binary tree where each node has either 0 or 2 children.
Problem
Approach
Steps
Complexity
Input: You are given a single integer n which represents the number of nodes in the full binary trees. n will always be an odd integer.
Example: Input: n = 5
Constraints:
• 1 <= n <= 20
• n is always an odd integer.
Output: Return a list of all possible full binary trees with n nodes. Each tree in the list should be represented by its root node, where each node has the value 0. The trees can be returned in any order.
Example: Output: [[0,0,0,null,null,0,0],[0,0,0,0,0,0,0]]
Constraints:
• The output should contain all unique full binary trees with n nodes.
Goal: The goal is to generate all unique full binary trees for a given number of nodes n.
Steps:
• 1. Use dynamic programming (DP) to generate all possible full binary trees with n nodes.
• 2. For each odd number of nodes i (1, 3, 5, ..., n), split the nodes into two subtrees by considering all possible left and right subtree sizes.
• 3. Recursively generate full binary trees for smaller subproblems and combine them into larger trees.
• 4. Store the root node of each tree and ensure the trees are unique.
Goal: The problem ensures that the number of nodes will be small enough (1 <= n <= 20) that dynamic programming can be applied efficiently.
Steps:
• The value of n is always an odd integer.
• 1 <= n <= 20.
Assumptions:
• The value of n will always be odd.
• Input: Input: n = 5
• Explanation: In this case, there are two possible full binary trees with 5 nodes. The trees are constructed by considering all possible ways to split the nodes into left and right subtrees.
• Input: Input: n = 7
• Explanation: For n = 7, there are several possible full binary trees. Each tree has 3 nodes in the left and right subtrees, with a root node connecting them.
Approach: The approach uses dynamic programming (DP) to build all possible full binary trees for n nodes. For each odd number n, recursively generate the left and right subtrees and combine them into full binary trees.
Observations:
• Each tree must have an odd number of nodes, as each non-leaf node must have exactly two children.
• Dynamic programming can be used to efficiently generate all possible full binary trees for smaller values of n and combine them.
• We can use a bottom-up approach, starting with small values of n and incrementally constructing larger trees.
Steps:
• 1. Create a DP array where dp[i] contains all possible full binary trees with i nodes.
• 2. Initialize dp[1] with a single node tree (the base case).
• 3. For each odd number i from 3 to n, generate full binary trees by combining smaller trees stored in dp.
• 4. Return dp[n] as the list of all possible full binary trees with n nodes.
Empty Inputs:
• If n = 1, the only possible tree is a single node.
Large Inputs:
• The solution should efficiently handle values of n up to 20.
Special Values:
• Since n is always an odd number, the solution can be designed to only process odd numbers.
Constraints:
• The solution should handle n up to 20 without performance issues.
vector<TreeNode*> allPossibleFBT(int n) {
vector<vector<TreeNode*>> dp(n + 1);
dp[1].push_back(new TreeNode(0));
for(int i = 3; i <= n; i += 2) {
for(int l = 1; l < i; l += 2) {
// cout << l;
for(auto it: dp[l])
for(auto it2: dp[i - l - 1]) {
TreeNode* node = new TreeNode(0);
node->left = it;
node->right = it2;
dp[i].push_back(node);
}
}
}
return dp[n];
}
1 : Function Definition
vector<TreeNode*> allPossibleFBT(int n) {
Define the function `allPossibleFBT`, which takes an integer `n` as input and returns a vector of pointers to TreeNode, representing all possible full binary trees with `n` nodes.
2 : Empty Line
This line is empty and used for formatting purposes.
3 : Dynamic Programming Table Initialization
vector<vector<TreeNode*>> dp(n + 1);
Initialize a 2D vector `dp` where `dp[i]` will store all possible full binary trees with `i` nodes.
4 : Base Case Initialization
dp[1].push_back(new TreeNode(0));
For the base case, initialize `dp[1]` with a single node tree. This is the only full binary tree with 1 node.
5 : Outer Loop (Iterating over odd numbers)
for(int i = 3; i <= n; i += 2) {
Start a loop from `i = 3` to `n`, incrementing by 2. This ensures that only odd values of `i` are considered, as full binary trees can only have an odd number of nodes.
6 : Inner Loop (Iterating over left subtree sizes)
for(int l = 1; l < i; l += 2) {
Start a loop to iterate over possible sizes of the left subtree, where `l` is the size of the left subtree, and it must be odd.
7 : Comment
// cout << l;
This line is a commented-out debug statement that would print the current size of the left subtree.
8 : Loop Through Left Subtree Trees
for(auto it: dp[l])
Iterate over all trees in `dp[l]`, which represent possible left subtrees for a tree with `i` nodes.
9 : Loop Through Right Subtree Trees
for(auto it2: dp[i - l - 1]) {
Iterate over all trees in `dp[i - l - 1]`, which represent possible right subtrees for a tree with `i` nodes.
10 : TreeNode Creation
TreeNode* node = new TreeNode(0);
Create a new tree node to serve as the root of the current full binary tree.
11 : Assign Left Subtree
node->left = it;
Assign the left child of the current node to be one of the trees from `dp[l]`.
12 : Assign Right Subtree
node->right = it2;
Assign the right child of the current node to be one of the trees from `dp[i - l - 1]`.
13 : Store Created Tree
dp[i].push_back(node);
Add the newly created full binary tree to `dp[i]`, which stores all possible full binary trees with `i` nodes.
14 : End Inner Loop
}
End the loop that iterates over possible right subtree trees.
15 : End Left Subtree Loop
}
End the loop that iterates over possible left subtree sizes.
16 : End Outer Loop
}
End the loop that iterates over all possible tree sizes `i`.
17 : Return Result
return dp[n];
Return the vector `dp[n]`, which contains all possible full binary trees with `n` nodes.
18 : End Function
}
End the function definition.
Best Case: O(n^2)
Average Case: O(n^3)
Worst Case: O(n^3)
Description: The time complexity is O(n^3) in the worst case, where n is the number of nodes. The dynamic programming approach generates and combines trees for each odd value of n.
Best Case: O(n)
Worst Case: O(n^2)
Description: The space complexity is O(n^2) in the worst case, where n is the number of nodes. This is due to the storage required for the DP array and the binary trees generated.
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