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894. 所有可能的真二叉树

English Version

题目描述

给你一个整数 n ，请你找出所有可能含 n 个节点的 真二叉树 ，并以列表形式返回。答案中每棵树的每个节点都必须符合 Node.val == 0 。

答案的每个元素都是一棵真二叉树的根节点。你可以按 任意顺序 返回最终的真二叉树列表。

真二叉树 是一类二叉树，树中每个节点恰好有 0 或 2 个子节点。

 

示例 1：

输入：n = 7
输出：[[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]]

示例 2：

输入：n = 3
输出：[[0,0,0]]

 

提示：

• 1 <= n <= 20

解法

方法一：记忆化搜索

如果 $n=1$，直接返回单个节点的列表。

如果 $n>1$，我们可以枚举左子树的节点数量 $i$，那么右子树的节点数量为 $n-1-i$。对于每种情况，我们递归地构造左子树和右子树的所有可能的真二叉树。然后将左子树和右子树两两组合，得到所有可能的真二叉树。

此过程可以用记忆化搜索，避免重复计算。

时间复杂度 $O(2^n)$，空间复杂度 $O(2^n)$。其中 $n$ 是节点数量。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def allPossibleFBT(self, n: int) -> List[Optional[TreeNode]]:
        @cache
        def dfs(n: int) -> List[Optional[TreeNode]]:
            if n == 1:
                return [TreeNode()]
            ans = []
            for i in range(n - 1):
                j = n - 1 - i
                for left in dfs(i):
                    for right in dfs(j):
                        ans.append(TreeNode(0, left, right))
            return ans

        return dfs(n)

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private List<TreeNode>[] f;

    public List<TreeNode> allPossibleFBT(int n) {
        f = new List[n + 1];
        return dfs(n);
    }

    private List<TreeNode> dfs(int n) {
        if (f[n] != null) {
            return f[n];
        }
        if (n == 1) {
            return List.of(new TreeNode());
        }
        List<TreeNode> ans = new ArrayList<>();
        for (int i = 0; i < n - 1; ++i) {
            int j = n - 1 - i;
            for (var left : dfs(i)) {
                for (var right : dfs(j)) {
                    ans.add(new TreeNode(0, left, right));
                }
            }
        }
        return f[n] = ans;
    }
}

C++

/**
 * 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:
    vector<TreeNode*> allPossibleFBT(int n) {
        vector<vector<TreeNode*>> f(n + 1);
        function<vector<TreeNode*>(int)> dfs = [&](int n) -> vector<TreeNode*> {
            if (f[n].size()) {
                return f[n];
            }
            if (n == 1) {
                return vector<TreeNode*>{new TreeNode()};
            }
            vector<TreeNode*> ans;
            for (int i = 0; i < n - 1; ++i) {
                int j = n - 1 - i;
                for (auto left : dfs(i)) {
                    for (auto right : dfs(j)) {
                        ans.push_back(new TreeNode(0, left, right));
                    }
                }
            }
            return f[n] = ans;
        };
        return dfs(n);
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func allPossibleFBT(n int) []*TreeNode {
	f := make([][]*TreeNode, n+1)
	var dfs func(int) []*TreeNode
	dfs = func(n int) []*TreeNode {
		if len(f[n]) > 0 {
			return f[n]
		}
		if n == 1 {
			return []*TreeNode{&TreeNode{Val: 0}}
		}
		ans := []*TreeNode{}
		for i := 0; i < n-1; i++ {
			j := n - 1 - i
			for _, left := range dfs(i) {
				for _, right := range dfs(j) {
					ans = append(ans, &TreeNode{0, left, right})
				}
			}
		}
		f[n] = ans
		return ans
	}
	return dfs(n)
}

TypeScript

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function allPossibleFBT(n: number): Array<TreeNode | null> {
    const f: Array<Array<TreeNode | null>> = new Array(n + 1)
        .fill(0)
        .map(() => []);
    const dfs = (n: number): Array<TreeNode | null> => {
        if (f[n].length) {
            return f[n];
        }
        if (n === 1) {
            f[n].push(new TreeNode(0));
            return f[n];
        }
        const ans: Array<TreeNode | null> = [];
        for (let i = 0; i < n - 1; ++i) {
            const j = n - 1 - i;
            for (const left of dfs(i)) {
                for (const right of dfs(j)) {
                    ans.push(new TreeNode(0, left, right));
                }
            }
        }
        return (f[n] = ans);
    };
    return dfs(n);
}

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