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feat: add solutions to lc/lcof2 problem:Sum Root to Leaf Numbers
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-96
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27 files changed

+537
-96
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‎lcof2/剑指 Offer II 049. 从根节点到叶节点的路径数字之和/README.md

Lines changed: 105 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -63,28 +63,126 @@
6363

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<!-- 这里可写通用的实现逻辑 -->
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DFS。
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<!-- tabs:start -->
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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 sumNumbers(self, root: TreeNode) -> int:
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def dfs(root, presum):
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if root is None:
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return 0
86+
s = 10 * presum + root.val
87+
if root.left is None and root.right is None:
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return s
89+
return dfs(root.left, s) + dfs(root.right, s)
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91+
return dfs(root, 0)
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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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### **...**
85-
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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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public int sumNumbers(TreeNode root) {
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return dfs(root, 0);
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}
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private int dfs(TreeNode root, int presum) {
120+
if (root == null) {
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return 0;
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}
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int s = presum * 10 + root.val;
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if (root.left == null && root.right == null) {
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return s;
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}
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return dfs(root.left, s) + dfs(root.right, s);
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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) {}
144+
* };
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*/
146+
class Solution {
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public:
148+
int sumNumbers(TreeNode *root) {
149+
return dfs(root, 0);
150+
}
151+
152+
int dfs(TreeNode *root, int presum) {
153+
if (root == nullptr)
154+
return 0;
155+
int s = presum * 10 + root->val;
156+
if (root->left == nullptr && root->right == nullptr)
157+
return s;
158+
return dfs(root->left, s) + dfs(root->right, s);
159+
}
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};
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```
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90-
<!-- tabs:end -->
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### **Go**
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```go
166+
/**
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* Definition for a binary tree node.
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* type TreeNode struct {
169+
* Val int
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* Left *TreeNode
171+
* Right *TreeNode
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* }
173+
*/
174+
func sumNumbers(root *TreeNode) int {
175+
return dfs(root, 0)
176+
}
177+
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func dfs(root *TreeNode, presum int) int {
179+
if root == nil {
180+
return 0
181+
}
182+
s := presum*10 + root.Val
183+
if root.Left == nil && root.Right == nil {
184+
return s
185+
}
186+
return dfs(root.Left, s) + dfs(root.Right, s)
187+
}
188+
```

‎lcof2/剑指 Offer II 049. 从根节点到叶节点的路径数字之和/Solution.cpp

Lines changed: 26 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,26 @@
1+
/**
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* Definition for a binary tree node.
3+
* struct TreeNode {
4+
* int val;
5+
* TreeNode *left;
6+
* TreeNode *right;
7+
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
8+
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
9+
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
10+
* };
11+
*/
12+
class Solution {
13+
public:
14+
int sumNumbers(TreeNode *root) {
15+
return dfs(root, 0);
16+
}
17+
18+
int dfs(TreeNode *root, int presum) {
19+
if (root == nullptr)
20+
return 0;
21+
int s = presum * 10 + root->val;
22+
if (root->left == nullptr && root->right == nullptr)
23+
return s;
24+
return dfs(root->left, s) + dfs(root->right, s);
25+
}
26+
};

‎lcof2/剑指 Offer II 049. 从根节点到叶节点的路径数字之和/Solution.go

Lines changed: 22 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,22 @@
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/**
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* Definition for a binary tree node.
3+
* type TreeNode struct {
4+
* Val int
5+
* Left *TreeNode
6+
* Right *TreeNode
7+
* }
8+
*/
9+
func sumNumbers(root *TreeNode) int {
10+
return dfs(root, 0)
11+
}
12+
13+
func dfs(root *TreeNode, presum int) int {
14+
if root == nil {
15+
return 0
16+
}
17+
s := presum*10 + root.Val
18+
if root.Left == nil && root.Right == nil {
19+
return s
20+
}
21+
return dfs(root.Left, s) + dfs(root.Right, s)
22+
}

‎lcof2/剑指 Offer II 049. 从根节点到叶节点的路径数字之和/Solution.java

Lines changed: 31 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,31 @@
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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;
5+
* TreeNode left;
6+
* TreeNode right;
7+
* TreeNode() {}
8+
* TreeNode(int val) { this.val = val; }
9+
* TreeNode(int val, TreeNode left, TreeNode right) {
10+
* this.val = val;
11+
* this.left = left;
12+
* this.right = right;
13+
* }
14+
* }
15+
*/
16+
class Solution {
17+
public int sumNumbers(TreeNode root) {
18+
return dfs(root, 0);
19+
}
20+
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private int dfs(TreeNode root, int presum) {
22+
if (root == null) {
23+
return 0;
24+
}
25+
int s = presum * 10 + root.val;
26+
if (root.left == null && root.right == null) {
27+
return s;
28+
}
29+
return dfs(root.left, s) + dfs(root.right, s);
30+
}
31+
}

‎lcof2/剑指 Offer II 049. 从根节点到叶节点的路径数字之和/Solution.py

Lines changed: 17 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,17 @@
1+
# Definition for a binary tree node.
2+
# class TreeNode:
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# def __init__(self, val=0, left=None, right=None):
4+
# self.val = val
5+
# self.left = left
6+
# self.right = right
7+
class Solution:
8+
def sumNumbers(self, root: TreeNode) -> int:
9+
def dfs(root, presum):
10+
if root is None:
11+
return 0
12+
s = 10 * presum + root.val
13+
if root.left is None and root.right is None:
14+
return s
15+
return dfs(root.left, s) + dfs(root.right, s)
16+
17+
return dfs(root, 0)

‎lcof2/剑指 Offer II 105. 岛屿的最大面积/README.md

Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -53,7 +53,7 @@ DFS 或并查集实现。
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模板 1——朴素并查集：
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```python
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# 初始化，p存储每个点的祖宗节点
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# 初始化，p存储每个点的父节点
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p = list(range(n))
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# 返回x的祖宗节点
@@ -70,7 +70,7 @@ p[find(a)] = find(b)
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模板 2——维护 size 的并查集：
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```python
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# 初始化，p存储每个点的祖宗节点，size只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
73+
# 初始化，p存储每个点的父节点，size只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
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p = list(range(n))
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size = [1] * n
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@@ -90,7 +90,7 @@ if find(a) != find(b):
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模板 3——维护到祖宗节点距离的并查集：
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```python
93-
# 初始化，p存储每个点的祖宗节点，d[x]存储x到p[x]的距离
93+
# 初始化，p存储每个点的父节点，d[x]存储x到p[x]的距离
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p = list(range(n))
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d = [0] * n
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‎lcof2/剑指 Offer II 116. 朋友圈/README.md

Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -60,7 +60,7 @@
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模板 1——朴素并查集：
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```python
63-
# 初始化，p存储每个点的祖宗节点
63+
# 初始化，p存储每个点的父节点
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p = list(range(n))
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# 返回x的祖宗节点
@@ -77,7 +77,7 @@ p[find(a)] = find(b)
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模板 2——维护 size 的并查集：
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```python
80-
# 初始化，p存储每个点的祖宗节点，size只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
80+
# 初始化，p存储每个点的父节点，size只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
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p = list(range(n))
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size = [1] * n
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@@ -97,7 +97,7 @@ if find(a) != find(b):
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模板 3——维护到祖宗节点距离的并查集：
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```python
100-
# 初始化，p存储每个点的祖宗节点，d[x]存储x到p[x]的距离
100+
# 初始化，p存储每个点的父节点，d[x]存储x到p[x]的距离
101101
p = list(range(n))
102102
d = [0] * n
103103

‎lcof2/剑指 Offer II 118. 多余的边/README.md

Lines changed: 3 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -57,7 +57,7 @@
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模板 1——朴素并查集：
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```python
60-
# 初始化，p存储每个点的祖宗节点
60+
# 初始化，p存储每个点的父节点
6161
p = list(range(n))
6262

6363
# 返回x的祖宗节点
@@ -74,7 +74,7 @@ p[find(a)] = find(b)
7474
模板 2——维护 size 的并查集：
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```python
77-
# 初始化，p存储每个点的祖宗节点，size只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
77+
# 初始化，p存储每个点的父节点，size只有当节点是祖宗节点时才有意义，表示祖宗节点所在集合中，点的数量
7878
p = list(range(n))
7979
size = [1] * n
8080

@@ -94,7 +94,7 @@ if find(a) != find(b):
9494
模板 3——维护到祖宗节点距离的并查集：
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9696
```python
97-
# 初始化，p存储每个点的祖宗节点，d[x]存储x到p[x]的距离
97+
# 初始化，p存储每个点的父节点，d[x]存储x到p[x]的距离
9898
p = list(range(n))
9999
d = [0] * n
100100

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