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Nov 6, 2024
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feat: add solutions to lc problem: No.2265 #3724

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113 changes: 77 additions & 36 deletions solution/2200-2299/2265.Count Nodes Equal to Average of Subtree/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -65,31 +65,36 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：DFS

我们设计一个函数 $\textit{dfs}$，它的作用是计算以当前节点为根的子树的和以及节点个数。

函数 $\textit{dfs}$ 的执行过程如下：

- 如果当前节点为空，返回 $(0, 0)$。
- 否则，我们递归计算左右子树的和以及节点个数，分别记为 $(\textit{ls}, \textit{ln})$ 和 $(\textit{rs}, \textit{rn})$。那么，以当前节点为根的子树的和 $\textit{s}$ 和节点个数 $\textit{n}$ 分别为 $\textit{ls} + \textit{rs} + \textit{root.val}$ 和 $\textit{ln} + \textit{rn} + 1$。如果 $\textit{s} / \textit{n} = \textit{root.val}$，则说明当前节点满足题目要求，我们将答案 $\textit{ans}$ 自增 $1$。
- 最后，函数 $\textit{dfs}$ 返回 $\textit{s}$ 和 $\textit{n}$。

我们初始化答案 $\textit{ans}$ 为 $0$，然后调用 $\textit{dfs}$ 函数，最后返回答案 $\textit{ans}$。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 表示二叉树的节点个数。

<!-- tabs:start -->

#### Python3

```python
# 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 averageOfSubtree(self, root: Optional[TreeNode]) -> int:
def dfs(root):
if root is None:
def averageOfSubtree(self, root: TreeNode) -> int:
def dfs(root) -> tuple:
if not root:
return 0, 0
ls, ln = dfs(root.left)
rs, rn = dfs(root.right)
s = ls + rs + root.val
n = ln + rn + 1
if s // n == root.val:
nonlocal ans
ans += 1
nonlocal ans
ans += int(s // n == root.val)
return s, n

ans = 0
Expand Down Expand Up @@ -119,17 +124,16 @@ class Solution {
private int ans;

public int averageOfSubtree(TreeNode root) {
ans = 0;
dfs(root);
return ans;
}

private int[] dfs(TreeNode root) {
if (root == null) {
return new int[] {0, 0};
return new int[2];
}
int[] l = dfs(root.left);
int[] r = dfs(root.right);
var l = dfs(root.left);
var r = dfs(root.right);
int s = l[0] + r[0] + root.val;
int n = l[1] + r[1] + 1;
if (s / n == root.val) {
Expand All @@ -156,22 +160,24 @@ class Solution {
*/
class Solution {
public:
int ans;
int averageOfSubtree(TreeNode* root) {
ans = 0;
dfs(root);
int ans = 0;
auto dfs = [&](auto&& dfs, TreeNode* root) -> pair<int, int> {
if (!root) {
return {0, 0};
}
auto [ls, ln] = dfs(dfs, root->left);
auto [rs, rn] = dfs(dfs, root->right);
int s = ls + rs + root->val;
int n = ln + rn + 1;
if (s / n == root->val) {
++ans;
}
return {s, n};
};
dfs(dfs, root);
return ans;
}

vector<int> dfs(TreeNode* root) {
if (!root) return {0, 0};
auto l = dfs(root->left);
auto r = dfs(root->right);
int s = l[0] + r[0] + root->val;
int n = l[1] + r[1] + 1;
if (s / n == root->val) ++ans;
return {s, n};
}
};
```

Expand All @@ -186,24 +192,59 @@ public:
* Right *TreeNode
* }
*/
func averageOfSubtree(root *TreeNode) int {
ans := 0
var dfs func(*TreeNode) (int, int)
func averageOfSubtree(root *TreeNode) (ans int) {
var dfs func(root *TreeNode) (int, int)
dfs = func(root *TreeNode) (int, int) {
if root == nil {
return 0, 0
}
ls, ln := dfs(root.Left)
rs, rn := dfs(root.Right)
s := ls + rs + root.Val
n := ln + rn + 1
s, n := ls+rs+root.Val, ln+rn+1
if s/n == root.Val {
ans++
}
return s, n
}
dfs(root)
return ans
return
}
```

#### TypeScript

```ts
/**
* 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 averageOfSubtree(root: TreeNode | null): number {
let ans: number = 0;
const dfs = (root: TreeNode | null): [number, number] => {
if (!root) {
return [0, 0];
}
const [ls, ln] = dfs(root.left);
const [rs, rn] = dfs(root.right);
const s = ls + rs + root.val;
const n = ln + rn + 1;
if (Math.floor(s / n) === root.val) {
++ans;
}
return [s, n];
};
dfs(root);
return ans;
}
```

Expand Down
115 changes: 78 additions & 37 deletions solution/2200-2299/2265.Count Nodes Equal to Average of Subtree/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -35,7 +35,7 @@ tags:
<pre>
<strong>Input:</strong> root = [4,8,5,0,1,null,6]
<strong>Output:</strong> 5
<strong>Explanation:</strong>
<strong>Explanation:</strong>
For the node with value 4: The average of its subtree is (4 + 8 + 5 + 0 + 1 + 6) / 6 = 24 / 6 = 4.
For the node with value 5: The average of its subtree is (5 + 6) / 2 = 11 / 2 = 5.
For the node with value 0: The average of its subtree is 0 / 1 = 0.
Expand Down Expand Up @@ -65,31 +65,36 @@ For the node with value 6: The average of its subtree is 6 / 1 = 6.

<!-- solution:start -->

### Solution 1
### Solution 1: DFS

We design a function $\textit{dfs}$, which calculates the sum and the number of nodes of the subtree rooted at the current node.

The execution process of the function $\textit{dfs}$ is as follows:

- If the current node is null, return $(0, 0)$.
- Otherwise, we recursively calculate the sum and the number of nodes of the left and right subtrees, denoted as $(\textit{ls}, \textit{ln})$ and $(\textit{rs}, \textit{rn})$, respectively. Then, the sum $\textit{s}$ and the number of nodes $\textit{n}$ of the subtree rooted at the current node are $\textit{ls} + \textit{rs} + \textit{root.val}$ and $\textit{ln} + \textit{rn} + 1$, respectively. If $\textit{s} / \textit{n} = \textit{root.val}$, it means the current node meets the requirement of the problem, and we increment the answer $\textit{ans}$ by $1$.
- Finally, the function $\textit{dfs}$ returns $\textit{s}$ and $\textit{n}$.

We initialize the answer $\textit{ans}$ to $0$, then call the $\textit{dfs}$ function, and finally return the answer $\textit{ans}$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ represents the number of nodes in the binary tree.

<!-- tabs:start -->

#### Python3

```python
# 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 averageOfSubtree(self, root: Optional[TreeNode]) -> int:
def dfs(root):
if root is None:
def averageOfSubtree(self, root: TreeNode) -> int:
def dfs(root) -> tuple:
if not root:
return 0, 0
ls, ln = dfs(root.left)
rs, rn = dfs(root.right)
s = ls + rs + root.val
n = ln + rn + 1
if s // n == root.val:
nonlocal ans
ans += 1
nonlocal ans
ans += int(s // n == root.val)
return s, n

ans = 0
Expand Down Expand Up @@ -119,17 +124,16 @@ class Solution {
private int ans;

public int averageOfSubtree(TreeNode root) {
ans = 0;
dfs(root);
return ans;
}

private int[] dfs(TreeNode root) {
if (root == null) {
return new int[] {0, 0};
return new int[2];
}
int[] l = dfs(root.left);
int[] r = dfs(root.right);
var l = dfs(root.left);
var r = dfs(root.right);
int s = l[0] + r[0] + root.val;
int n = l[1] + r[1] + 1;
if (s / n == root.val) {
Expand All @@ -156,22 +160,24 @@ class Solution {
*/
class Solution {
public:
int ans;
int averageOfSubtree(TreeNode* root) {
ans = 0;
dfs(root);
int ans = 0;
auto dfs = [&](auto&& dfs, TreeNode* root) -> pair<int, int> {
if (!root) {
return {0, 0};
}
auto [ls, ln] = dfs(dfs, root->left);
auto [rs, rn] = dfs(dfs, root->right);
int s = ls + rs + root->val;
int n = ln + rn + 1;
if (s / n == root->val) {
++ans;
}
return {s, n};
};
dfs(dfs, root);
return ans;
}

vector<int> dfs(TreeNode* root) {
if (!root) return {0, 0};
auto l = dfs(root->left);
auto r = dfs(root->right);
int s = l[0] + r[0] + root->val;
int n = l[1] + r[1] + 1;
if (s / n == root->val) ++ans;
return {s, n};
}
};
```

Expand All @@ -186,24 +192,59 @@ public:
* Right *TreeNode
* }
*/
func averageOfSubtree(root *TreeNode) int {
ans := 0
var dfs func(*TreeNode) (int, int)
func averageOfSubtree(root *TreeNode) (ans int) {
var dfs func(root *TreeNode) (int, int)
dfs = func(root *TreeNode) (int, int) {
if root == nil {
return 0, 0
}
ls, ln := dfs(root.Left)
rs, rn := dfs(root.Right)
s := ls + rs + root.Val
n := ln + rn + 1
s, n := ls+rs+root.Val, ln+rn+1
if s/n == root.Val {
ans++
}
return s, n
}
dfs(root)
return ans
return
}
```

#### TypeScript

```ts
/**
* 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 averageOfSubtree(root: TreeNode | null): number {
let ans: number = 0;
const dfs = (root: TreeNode | null): [number, number] => {
if (!root) {
return [0, 0];
}
const [ls, ln] = dfs(root.left);
const [rs, rn] = dfs(root.right);
const s = ls + rs + root.val;
const n = ln + rn + 1;
if (Math.floor(s / n) === root.val) {
++ans;
}
return [s, n];
};
dfs(root);
return ans;
}
```

Expand Down
30 changes: 16 additions & 14 deletions solution/2200-2299/2265.Count Nodes Equal to Average of Subtree/Solution.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -11,20 +11,22 @@
*/
class Solution {
public:
int ans;
int averageOfSubtree(TreeNode* root) {
ans = 0;
dfs(root);
int ans = 0;
auto dfs = [&](auto&& dfs, TreeNode* root) -> pair<int, int> {
if (!root) {
return {0, 0};
}
auto [ls, ln] = dfs(dfs, root->left);
auto [rs, rn] = dfs(dfs, root->right);
int s = ls + rs + root->val;
int n = ln + rn + 1;
if (s / n == root->val) {
++ans;
}
return {s, n};
};
dfs(dfs, root);
return ans;
}

vector<int> dfs(TreeNode* root) {
if (!root) return {0, 0};
auto l = dfs(root->left);
auto r = dfs(root->right);
int s = l[0] + r[0] + root->val;
int n = l[1] + r[1] + 1;
if (s / n == root->val) ++ans;
return {s, n};
}
};
};
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