根据节点数目构建满二叉树

894. All Possible Full Binary Tree

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

Return a list of all possible full binary trees with N nodes.  Each element of the answer is the root node of one possible tree.

Each node of each tree in the answer must have node.val = 0.

You may return the final list of trees in any order.

 

Example 1:

Input: 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]]
Explanation:

 

Note:

• 1 <= N <= 20

题解：从底层往顶层构建，例如当前须要构建节点数为5的满二叉树，那么可行的划分为：左子树1个，右子树3个；或者左子树3个，右子树1个；2-2的划分不合法。那么，当底层已经构建完成时，只须要枚举左、右子树的组合便可。

class Solution {
public:
    vector<TreeNode*> allPossibleFBT(int N) {
        vector<TreeNode*>ans;
        if(N==1){
            ans.push_back(new TreeNode(0));
            return ans;
        }
        if(N%2==0) return ans;  //偶数节点不能构成满二叉树
        for(int i=1;i<N;i+=2){ //N=根节点+左子树为i个节点，右子树为N-1-i个节点，
            if(N-1-i<1) continue;
            auto left=allPossibleFBT(i);
            auto right=allPossibleFBT(N-1-i);
            for(auto&l:left){
                for(auto&r:right){
                    auto tmp=new TreeNode(0);
                    tmp->left=l;
                    tmp->right=r;
                    ans.push_back(tmp);
                }
            }
        }
        return ans;
    }
};