894. All Possible Full Binary Trees

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894. All Possible Full Binary Trees

Description

Given an integer n, return a list of all possible full binary trees with n nodes. Each node of each tree in the answer must have Node.val == 0.

Each element of the answer is the root node of one possible tree. You may return the final list of trees in any order .

A full binary tree is a binary tree where each node has exactly 0 or 2 children.

Example 1:

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Input: n = 7
Output: [[0,0,0,null,null,0,0,null,null,0,0],[0,0,0,null,null,0,0,0,0],[0,0,0,0,0,0,0],[0,0,0,0,0,null,null,null,null,0,0],[0,0,0,0,0,null,null,0,0]]

Example 2:

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Input: n = 3
Output: [[0,0,0]]

Constraints:

  • 1 <= n <= 20

Hints/Notes

  • N/A

Solution

Language: C++

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    map<int, vector<TreeNode*>> m;

    vector<TreeNode*> allPossibleFBT(int n) {
        vector<TreeNode*> res;
        if (n % 2 == 0) {
            return res;
        }
        return build(n);
    }

    vector<TreeNode*> build(int n) {
        if (m.contains(n)) {
            return m[n];
        }
        vector<TreeNode*> res;
        if (n == 1) {
            res.push_back(new TreeNode(0));
            m[n] = res;
            return res;
        }
        for (int i = 1; i < n; i += 2) {
            int j = n - i - 1;
            auto left = build(i);
            auto right = build(j);
            for (auto l : left) {
                for (auto r : right) {
                    TreeNode* root = new TreeNode(0);
                    root->left = l;
                    root->right = r;
                    res.push_back(root);
                }
            }
        }
        m[n] = res;
        return res;
    }
};