All Possible Full Binary Trees

class Solution {
public:
    vector<TreeNode*> generateAllFBT(int totalNodes) {
        if (totalNodes % 2 == 0) {
            return {};
        }
        
        vector<vector<TreeNode*>> trees(totalNodes + 1);
        trees[1].push_back(new TreeNode(0));
        
        for (int nodes = 3; nodes <= totalNodes; nodes += 2) {
            for (int leftNodes = 1; leftNodes < nodes - 1; leftNodes += 2) {
                int rightNodes = nodes - 1 - leftNodes;
                for (auto leftTree : trees[leftNodes]) {
                    for (auto rightTree : trees[rightNodes]) {
                        TreeNode* currentNode = new TreeNode(0, leftTree, rightTree);
                        trees[nodes].push_back(currentNode);
                    }
                }
            }
        }
        
        return trees[totalNodes];
    }
};

The provided C++ solution addresses the problem of generating all possible full binary trees given a specific number of nodes. Full binary trees are defined as trees where every node except the leaf nodes has exactly two child nodes. The implementation leverages dynamic programming to build trees of increasing sizes.

Key points in the solution:

• A full binary tree can only have an odd number of nodes, so the function quickly returns an empty vector if the total number of nodes is even.

1. Initialize a 2D vector called trees where each index represents the number of nodes and stores the possible full binary trees with that node count.
2. Hard code the base case for a single node tree.
3. Iterate over odd numbers from 3 up to totalNodes. For each number of nodes, break it down into left and right subtrees, considering all possible subtree combinations that sum up to nodes.
4. For each possible combination of left and right subtree node counts:
   • Iterate through all trees possible with leftNodes, use each as a left subtree.
   • Iterate through all trees possible with rightNodes, using each as a right subtree.
   • Create a new tree node with the left and right subtrees attached and push it into the trees vector at the current nodes position.

At the end of these steps, trees[totalNodes] will contain all possible full binary trees with the given number of nodes, and this vector is returned.

This approach is efficient by recycling previously computed results for smaller numbers of nodes and builds up to solve for the desired number of nodes, ensuring it only considers valid configurations by maintaining the structure of a full binary tree.

class Solution {
    public List<TreeNode> generateFBTs(int totalNodes) {
        if (totalNodes % 2 == 0) {
            return new ArrayList<>();
        }

        List<List<TreeNode>> memo = new ArrayList<>();
        for (int i = 0; i <= totalNodes; i++) {
            memo.add(new ArrayList<>());
        }

        memo.get(1).add(new TreeNode(0));
        for (int nodes = 3; nodes <= totalNodes; nodes += 2) {
            for (int leftCount = 1; leftCount < nodes - 1; leftCount += 2) {
                int rightCount = nodes - 1 - leftCount;
                for (TreeNode leftSubtree : memo.get(leftCount)) {
                    for (TreeNode rightSubtree : memo.get(rightCount)) {
                        TreeNode rootNode = new TreeNode(0, leftSubtree, rightSubtree);
                        memo.get(nodes).add(rootNode);
                    }
                }
            }
        }

        return memo.get(totalNodes);
    }
}

The provided Java solution is designed to generate all possible full binary trees with a specified number of nodes. Full binary trees are trees where every node other than the leaves has two children. The function generateFBTs employs a dynamic programming approach to solve this problem efficiently. Here is how the algorithm works:

1. First, the function checks if the totalNodes is even. A full binary tree cannot have an even number of nodes, so it returns an empty list if totalNodes is even.

2. Initialize memo as an ArrayList of ArrayLists of TreeNode. The memo list stores the list of all possible full binary trees with i nodes, where i ranges from 0 to totalNodes.

3. Seed the dynamic programming array: Add a base case to memo where the only possible tree with one node is a single TreeNode with no children.

4. Use a for-loop where nodes starts from 3 and goes up to totalNodes, incrementing by 2 at each step (since only odd numbers are considered).

5. For each value of nodes, compute trees by iterating over all possible counts of left and right children nodes, restricting these counts to odd numbers (since both left and right subtrees themselves also need to be full).

6. In the nested loops for each possible division of nodes between the left and right subtrees, create new trees by combining trees from memo.get(leftCount) and memo.get(rightCount) with a new root node.

7. Each combined tree forms a valid full binary tree for the current number of nodes, and is added to memo.get(nodes).

8. Return the list of all possible full binary trees with totalNodes from memo.

The core of this solution leverages memoization to avoid redundant calculations and efficiently generate large trees by building on smaller subtree solutions. The logical structuring of nodes, coupled with the dynamic use of memoization, ensures that all combinations of valid trees are explored and generated.

class BinaryTreeNode:
    def __init__(self, value=0, leftChild=None, rightChild=None):
        self.value = value
        self.leftChild = leftChild
        self.rightChild = rightChild

class Solution:
    def generateAllFBT(self, totalNodes: int) -> List[BinaryTreeNode]:
        if totalNodes % 2 == 0:
            return []

        memo = [[] for _ in range(totalNodes + 1)]
        memo[1].append(BinaryTreeNode(0))
        
        for nodes in range(3, totalNodes + 1, 2):
            for leftNodes in range(1, nodes - 1, 2):
                rightNodes = nodes - 1 - leftNodes
                for leftSubtree in memo[leftNodes]:
                    for rightSubtree in memo[rightNodes]:
                        parentNode = BinaryTreeNode(0, leftSubtree, rightSubtree)
                        memo[nodes].append(parentNode)

        return memo[totalNodes]

The presented Python 3 code defines a solution for generating all possible full binary trees with a given number of nodes. A full binary tree is characterized where every node, other than the leaves, has two children.

The solution class, named Solution, contains a method generateAllFBT which takes an integer totalNodes representing the total number of nodes in the tree. Here are the key steps used in the code:

1. The method checks if the total number of nodes is even. If true, it immediately returns an empty list as it's not possible to form a full binary tree with an even number of nodes.
2. Use memoization to store previously computed values in a list named memo. This list holds subtrees indexed by the number of nodes they contain.
3. Initialize memo[1] with a singleton list containing a single-node tree. This will serve as the base case.
4. Iterate over tree sizes starting from 3 to totalNodes, in steps of 2 (since only odd numbers are valid).
5. For each possible size (nodes), iterate through possible numbers of left subtree nodes (leftNodes).
   • Calculate rightNodes as nodes - 1 - leftNodes to maintain the full binary tree property.
   • For each left subtree combination in memo[leftNodes] and each right subtree combination in memo[rightNodes], create a new tree with these subtrees as children of a new root node and store this new tree in memo[nodes].
6. Return the list of all possible full binary trees with totalNodes from memo[totalNodes].

This approach cleverly uses dynamic programming to efficiently build trees by reusing previously calculated subtrees, reducing redundant calculations, thus optimizing the solution. The storage structure (memo) maps perfectly to the nature of the problem, making the solution elegant and effective.