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:
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:
Input: n = 3
Output: [[0,0,0]]
Constraints:
Solutions
Solution 1: Memoization Search
If n=1, return a list with a single node directly.
If n > 1, we can enumerate the number of nodes i in the left subtree, then the number of nodes in the right subtree is n-1-i. For each case, we recursively construct all possible genuine binary trees for the left and right subtrees. Then we combine the left and right subtrees in pairs to get all possible genuine binary trees.
This process can be optimized with memoization search to avoid repeated calculations.
The time complexity is O(2n⁄√n), and the space complexity is O(2n⁄√n). Where n is the number of nodes.
PythonJavaC++GoTypeScriptRustC#
# 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)(code-box)
/**
* 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;
}
}(code-box)
/**
* 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);
}
};(code-box)
/**
* 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)
}(code-box)
/**
* 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);
}(code-box)
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
// pub val: i32,
// pub left: Option<Rc<RefCell<TreeNode>>>,
// pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
// #[inline]
// pub fn new(val: i32) -> Self {
// TreeNode {
// val,
// left: None,
// right: None
// }
// }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
pub fn all_possible_fbt(n: i32) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
let mut f: Vec<Option<Vec<Option<Rc<RefCell<TreeNode>>>>>> = vec![None; (n + 1) as usize];
Self::dfs(n, &mut f)
}
fn dfs(
n: i32,
f: &mut Vec<Option<Vec<Option<Rc<RefCell<TreeNode>>>>>>,
) -> Vec<Option<Rc<RefCell<TreeNode>>>> {
if let Some(ref result) = f[n as usize] {
return result.clone();
}
let mut ans = Vec::new();
if n == 1 {
ans.push(Some(Rc::new(RefCell::new(TreeNode::new(0)))));
return ans;
}
for i in 0..n - 1 {
let j = n - 1 - i;
for left in Self::dfs(i, f).iter() {
for right in Self::dfs(j, f).iter() {
let new_node = Some(Rc::new(RefCell::new(TreeNode {
val: 0,
left: left.clone(),
right: right.clone(),
})));
ans.push(new_node);
}
}
}
f[n as usize] = Some(ans.clone());
ans
}
}(code-box)
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
private List<TreeNode>[] f;
public IList<TreeNode> AllPossibleFBT(int n) {
f = new List<TreeNode>[n + 1];
return Dfs(n);
}
private IList<TreeNode> Dfs(int n) {
if (f[n] != null) {
return f[n];
}
if (n == 1) {
return new List<TreeNode> { new TreeNode() };
}
List<TreeNode> ans = new List<TreeNode>();
for (int i = 0; i < n - 1; ++i) {
int j = n - 1 - i;
foreach (var left in Dfs(i)) {
foreach (var right in Dfs(j)) {
ans.Add(new TreeNode(0, left, right));
}
}
}
f[n] = ans;
return ans;
}
}(code-box)
