LeetCode-All Possible Full Binary Trees

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/Apple_hzc/article/details/83510572

一、Description

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.

题目大意：

给定一个数N，找出所有结点数为N的满数。满树指的是每个结点的子节点数要么为0要么为2。

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:

---

二、Analyzation

首先，对于N为偶数，不可能存在一个满足条件的满树，因此N必须为奇数。除开根结点，一棵满树可以分解为有i（i为奇数）个结点的左子树和n-i-1（n-i-1为奇数）个结点的右子树，所以只需遍历i的取值。

---

三、Accepted code

class Solution {
    public List<TreeNode> allPossibleFBT(int N) {
        List<TreeNode> list = new ArrayList<>();
        if (N % 2 == 0 || N <= 0) {
            return list;
        }
        if (N == 1) {
            list.add(new TreeNode(0));
            return list;
        }
        for (int i = 1; i < N; i += 2) {
            List<TreeNode> left = allPossibleFBT(i);
            List<TreeNode> right = allPossibleFBT(N - i - 1);
            for (TreeNode l : left) {
                for (TreeNode r : right) {
                    TreeNode node = new TreeNode(0);
                    node.left = l;
                    node.right = r;
                    list.add(node);
                }
            }
        }
        return list;
    }
}