LeetCode之All Possible Full Binar

问题：
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

方法：
类似斐波那契数列的原理，先生成1个node的tree，再由1个node的tree生成3个node的tree，直到迭代到N个node的tree即为最终结果，计算过程中通过HashMap保存生成的tree。

具体实现：

class AllPossibleFullBinaryTrees {
    // Definition for a binary tree node.
    class TreeNode(var `val`: Int = 0) {
        var left: TreeNode? = null
        var right: TreeNode? = null
    }

    private val map = mutableMapOf<Int, List<TreeNode?>>(1 to listOf(TreeNode(0)))

    fun allPossibleFBT(N: Int): List<TreeNode?> {
        if (!map.containsKey(N)) {
            val ans = mutableListOf<TreeNode?>()
            for (x in 0 until N) {
                val y = N - 1 - x
                for (left in allPossibleFBT(x)) {
                    for (right in allPossibleFBT(y)) {
                        val bns = TreeNode(0)
                        bns.left = left
                        bns.right = right
                        ans.add(bns)
                    }
                }
            }
            map.put(N, ans)
        }
        return map[N].orEmpty()
    }
}

fun main(args: Array<String>) {

}

有问题随时沟通

具体代码实现可以参考Github