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feat: add solutions to lc problem: No.0894
No.0894.All Possible Full Binary Trees
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‎solution/0800-0899/0894.All Possible Full Binary Trees/README.md

Lines changed: 149 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -40,22 +40,169 @@
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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**方法一：记忆化搜索**
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对于 $n=1$，直接返回单个节点的列表。
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对于 $n\gt1$，若 $n$ 为奇数，按照左右子树遍历不同的排列组合，得到结果列表；若 $n$ 为偶数，返回空列表。
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此过程可以用记忆化搜索，避免重复计算。
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时间复杂度 $O(2^n)$，空间复杂度 $O(2^n)$。
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
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# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
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def allPossibleFBT(self, n: int) -> List[Optional[TreeNode]]:
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@cache
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def dfs(n):
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if n == 1:
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return [TreeNode()]
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res = []
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if n % 2:
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for i in range(n - 1):
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j = n - i - 1
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for left in dfs(i):
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for right in dfs(j):
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res.append(TreeNode(0, left, right))
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return res
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return dfs(n)
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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class Solution {
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private List<TreeNode>[] f = new List[21];
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public List<TreeNode> allPossibleFBT(int n) {
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return dfs(n);
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}
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private List<TreeNode> dfs(int n) {
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if (f[n] != null) {
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return f[n];
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}
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if (n == 1) {
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return Collections.singletonList(new TreeNode());
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}
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List<TreeNode> res = new ArrayList<>();
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for (int i = 0; i < n - 1; ++i) {
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int j = n - i - 1;
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for (TreeNode left : dfs(i)) {
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for (TreeNode right : dfs(j)) {
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res.add(new TreeNode(0, left, right));
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}
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}
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}
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f[n] = res;
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return res;
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}
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}
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```
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### **C++**
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```cpp
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
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* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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* };
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*/
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class Solution {
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public:
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unordered_map<int, vector<TreeNode*>> f;
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vector<TreeNode*> allPossibleFBT(int n) {
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return dfs(n);
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}
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vector<TreeNode*> dfs(int n) {
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if (f.count(n)) return f[n];
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if (n == 1) return {new TreeNode()};
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vector<TreeNode*> res;
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for (int i = 0; i < n - 1; ++i) {
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int j = n - i - 1;
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for (auto left : dfs(i)) {
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for (auto right : dfs(j)) {
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res.push_back(new TreeNode(0, left, right));
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}
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}
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}
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f[n] = res;
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return res;
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}
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};
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```
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### **Go**
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```go
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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func allPossibleFBT(n int) []*TreeNode {
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f := map[int][]*TreeNode{}
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var dfs func(n int) []*TreeNode
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dfs = func(n int) []*TreeNode {
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if v, ok := f[n]; ok {
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return v
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}
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if n == 1 {
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return []*TreeNode{&TreeNode{Val: 0}}
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}
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res := []*TreeNode{}
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for i := 0; i < n-1; i++ {
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j := n - i - 1
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for _, left := range dfs(i) {
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for _, right := range dfs(j) {
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res = append(res, &TreeNode{0, left, right})
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}
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}
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}
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f[n] = res
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return res
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}
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return dfs(n)
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}
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```
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### **...**

‎solution/0800-0899/0894.All Possible Full Binary Trees/README_EN.md

Lines changed: 140 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -39,13 +39,152 @@
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### **Python3**
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```python
42-
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
45+
# self.val = val
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# self.left = left
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# self.right = right
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class Solution:
49+
def allPossibleFBT(self, n: int) -> List[Optional[TreeNode]]:
50+
@cache
51+
def dfs(n):
52+
if n == 1:
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return [TreeNode()]
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res = []
55+
if n % 2:
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for i in range(n - 1):
57+
j = n - i - 1
58+
for left in dfs(i):
59+
for right in dfs(j):
60+
res.append(TreeNode(0, left, right))
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return res
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return dfs(n)
4364
```
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### **Java**
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```java
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
73+
* TreeNode left;
74+
* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
82+
* }
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*/
84+
class Solution {
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private List<TreeNode>[] f = new List[21];
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public List<TreeNode> allPossibleFBT(int n) {
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return dfs(n);
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}
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private List<TreeNode> dfs(int n) {
92+
if (f[n] != null) {
93+
return f[n];
94+
}
95+
if (n == 1) {
96+
return Collections.singletonList(new TreeNode());
97+
}
98+
List<TreeNode> res = new ArrayList<>();
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for (int i = 0; i < n - 1; ++i) {
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int j = n - i - 1;
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for (TreeNode left : dfs(i)) {
102+
for (TreeNode right : dfs(j)) {
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res.add(new TreeNode(0, left, right));
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}
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}
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}
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f[n] = res;
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return res;
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}
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}
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```
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### **C++**
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```cpp
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/**
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* Definition for a binary tree node.
118+
* struct TreeNode {
119+
* int val;
120+
* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
125+
* };
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*/
127+
class Solution {
128+
public:
129+
unordered_map<int, vector<TreeNode*>> f;
130+
131+
vector<TreeNode*> allPossibleFBT(int n) {
132+
return dfs(n);
133+
}
134+
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vector<TreeNode*> dfs(int n) {
136+
if (f.count(n)) return f[n];
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if (n == 1) return {new TreeNode()};
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vector<TreeNode*> res;
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for (int i = 0; i < n - 1; ++i) {
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int j = n - i - 1;
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for (auto left : dfs(i)) {
142+
for (auto right : dfs(j)) {
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res.push_back(new TreeNode(0, left, right));
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}
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}
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}
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f[n] = res;
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return res;
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}
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};
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```
48152
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### **Go**
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```go
156+
/**
157+
* Definition for a binary tree node.
158+
* type TreeNode struct {
159+
* Val int
160+
* Left *TreeNode
161+
* Right *TreeNode
162+
* }
163+
*/
164+
func allPossibleFBT(n int) []*TreeNode {
165+
f := map[int][]*TreeNode{}
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var dfs func(n int) []*TreeNode
167+
dfs = func(n int) []*TreeNode {
168+
if v, ok := f[n]; ok {
169+
return v
170+
}
171+
if n == 1 {
172+
return []*TreeNode{&TreeNode{Val: 0}}
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}
174+
res := []*TreeNode{}
175+
for i := 0; i < n-1; i++ {
176+
j := n - i - 1
177+
for _, left := range dfs(i) {
178+
for _, right := range dfs(j) {
179+
res = append(res, &TreeNode{0, left, right})
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}
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}
182+
}
183+
f[n] = res
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return res
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}
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return dfs(n)
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}
49188
```
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### **...**

‎solution/0800-0899/0894.All Possible Full Binary Trees/Solution.cpp

Lines changed: 35 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,35 @@
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/**
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* Definition for a binary tree node.
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* struct TreeNode {
4+
* int val;
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* TreeNode *left;
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* TreeNode *right;
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* TreeNode() : val(0), left(nullptr), right(nullptr) {}
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* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
10+
* };
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*/
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class Solution {
13+
public:
14+
unordered_map<int, vector<TreeNode*>> f;
15+
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vector<TreeNode*> allPossibleFBT(int n) {
17+
return dfs(n);
18+
}
19+
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vector<TreeNode*> dfs(int n) {
21+
if (f.count(n)) return f[n];
22+
if (n == 1) return {new TreeNode()};
23+
vector<TreeNode*> res;
24+
for (int i = 0; i < n - 1; ++i) {
25+
int j = n - i - 1;
26+
for (auto left : dfs(i)) {
27+
for (auto right : dfs(j)) {
28+
res.push_back(new TreeNode(0, left, right));
29+
}
30+
}
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}
32+
f[n] = res;
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return res;
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}
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};

‎solution/0800-0899/0894.All Possible Full Binary Trees/Solution.go

Lines changed: 32 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,32 @@
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
4+
* Val int
5+
* Left *TreeNode
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* Right *TreeNode
7+
* }
8+
*/
9+
func allPossibleFBT(n int) []*TreeNode {
10+
f := map[int][]*TreeNode{}
11+
var dfs func(n int) []*TreeNode
12+
dfs = func(n int) []*TreeNode {
13+
if v, ok := f[n]; ok {
14+
return v
15+
}
16+
if n == 1 {
17+
return []*TreeNode{&TreeNode{Val: 0}}
18+
}
19+
res := []*TreeNode{}
20+
for i := 0; i < n-1; i++ {
21+
j := n - i - 1
22+
for _, left := range dfs(i) {
23+
for _, right := range dfs(j) {
24+
res = append(res, &TreeNode{0, left, right})
25+
}
26+
}
27+
}
28+
f[n] = res
29+
return res
30+
}
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return dfs(n)
32+
}

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