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feat: add solutions to lc problem: No.1022
No.1022.Sum of Root To Leaf Binary Numbers
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‎README.md

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## 贡献者
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非常感谢以下所有朋友对本项目的贡献，你们是最可爱的人！
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感谢以下所有朋友对本项目的贡献！
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<a href="https://github.com/doocs/leetcode/graphs/contributors" target="_blank"><img src="./images/contributors.svg" /></a>
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## 赞助者
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特别感谢以下个人、组织对本项目的赞助！
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感谢以下个人、组织对本项目的赞助！
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<a href="https://opencollective.com/doocs-leetcode/backers.svg?width=890" target="_blank"><img src="https://opencollective.com/doocs-leetcode/backers.svg?width=890"></a>
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‎lcci/README.md

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## 题解
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以下所有题目均来源 [LeetCode 中国官网](https://leetcode-cn.com/problemset/lcci/)，题解由 [doocs/leetcode 贡献者](https://github.com/doocs/leetcode/graphs/contributors) 提供，正在完善中，欢迎贡献你的题解！
9+
列表所有题解均由 [开源社区 Doocs](https://github.com/doocs) 贡献者提供，正在完善中，欢迎贡献你的题解！
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快速搜索题号、题解、标签等，请善用 <kbd>Control</kbd>+<kbd>F</kbd>（或者 <kbd>Command</kbd>+<kbd>F</kbd>）。
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‎lcof/README.md

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## 题解
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以下所有题目均来源 [LeetCode 中国官网](https://leetcode-cn.com/problemset/lcof/)，题解由 [doocs/leetcode 贡献者](https://github.com/doocs/leetcode/graphs/contributors) 提供，正在完善中，欢迎贡献你的题解！
11+
列表所有题解均由 [开源社区 Doocs](https://github.com/doocs) 贡献者提供，正在完善中，欢迎贡献你的题解！
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快速搜索题号、题解、标签等，请善用 <kbd>Control</kbd>+<kbd>F</kbd>（或者 <kbd>Command</kbd>+<kbd>F</kbd>）。
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‎lcof2/README.md

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## 题解
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以下所有题目均来源 [LeetCode 中国官网](https://leetcode-cn.com/problem-list/e8X3pBZi)，题解由 [doocs/leetcode 贡献者](https://github.com/doocs/leetcode/graphs/contributors) 提供，正在完善中，欢迎贡献你的题解！
9+
列表所有题解均由 [开源社区 Doocs](https://github.com/doocs) 贡献者提供，正在完善中，欢迎贡献你的题解！
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快速搜索题号、题解、标签等，请善用 <kbd>Control</kbd>+<kbd>F</kbd>（或者 <kbd>Command</kbd>+<kbd>F</kbd>）。
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‎lcp/README.md

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## 题解
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以下所有题目均来源 [LeetCode 中国官网](https://leetcode-cn.com)，题解由 [doocs/leetcode 贡献者](https://github.com/doocs/leetcode/graphs/contributors) 提供，正在完善中，欢迎贡献你的题解！
5+
列表所有题解均由 [开源社区 Doocs](https://github.com/doocs) 贡献者提供，正在完善中，欢迎贡献你的题解！
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快速搜索题号、题解、标签等，请善用 <kbd>Control</kbd>+<kbd>F</kbd>（或者 <kbd>Command</kbd>+<kbd>F</kbd>）。
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‎lcs/README.md

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## 题解
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以下所有题目均来源 [LeetCode 中国官网](https://leetcode-cn.com)，题解由 [doocs/leetcode 贡献者](https://github.com/doocs/leetcode/graphs/contributors) 提供，正在完善中，欢迎贡献你的题解！
5+
列表所有题解均由 [开源社区 Doocs](https://github.com/doocs) 贡献者提供，正在完善中，欢迎贡献你的题解！
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快速搜索题号、题解、标签等，请善用 <kbd>Control</kbd>+<kbd>F</kbd>（或者 <kbd>Command</kbd>+<kbd>F</kbd>）。
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‎solution/1000-1099/1022.Sum of Root To Leaf Binary Numbers/README.md

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<li><code>Node.val</code> 为 <code>0</code> 或 <code>1</code> 。</li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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深度优先搜索 DFS 实现。
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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 sumRootToLeaf(self, root: TreeNode) -> int:
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def dfs(root, t):
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if root is None:
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return 0
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t = (t << 1) | root.val
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if root.left is None and root.right is None:
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return t
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return dfs(root.left, t) + dfs(root.right, t)
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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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* 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 sumRootToLeaf(TreeNode root) {
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return dfs(root, 0);
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}
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private int dfs(TreeNode root, int t) {
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if (root == null) {
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return 0;
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}
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t = (t << 1) | root.val;
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if (root.left == null && root.right == null) {
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return t;
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}
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return dfs(root.left, t) + dfs(root.right, t);
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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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int sumRootToLeaf(TreeNode* root) {
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return dfs(root, 0);
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}
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int dfs(TreeNode* root, int t) {
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if (!root) return 0;
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t = (t << 1) | root->val;
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if (!root->left && !root->right) return t;
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return dfs(root->left, t) + dfs(root->right, t);
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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 sumRootToLeaf(root *TreeNode) int {
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var dfs func(root *TreeNode, t int) int
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dfs = func(root *TreeNode, t int) int {
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if root == nil {
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return 0
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}
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t = (t << 1) | root.Val
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if root.Left == nil && root.Right == nil {
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return t
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}
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return dfs(root.Left, t) + dfs(root.Right, t)
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}
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return dfs(root, 0)
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}
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```
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### **...**

‎solution/1000-1099/1022.Sum of Root To Leaf Binary Numbers/README_EN.md

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<li><code>Node.val</code> is <code>0</code> or <code>1</code>.</li>
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</ul>
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## Solutions
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DFS.
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<!-- tabs:start -->
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### **Python3**
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```python
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# Definition for a binary tree node.
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# class TreeNode:
62+
# def __init__(self, val=0, left=None, right=None):
63+
# self.val = val
64+
# self.left = left
65+
# self.right = right
66+
class Solution:
67+
def sumRootToLeaf(self, root: TreeNode) -> int:
68+
def dfs(root, t):
69+
if root is None:
70+
return 0
71+
t = (t << 1) | root.val
72+
if root.left is None and root.right is None:
73+
return t
74+
return dfs(root.left, t) + dfs(root.right, t)
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return dfs(root, 0)
6077
```
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### **Java**
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```java
82+
/**
83+
* Definition for a binary tree node.
84+
* public class TreeNode {
85+
* int val;
86+
* TreeNode left;
87+
* TreeNode right;
88+
* TreeNode() {}
89+
* TreeNode(int val) { this.val = val; }
90+
* TreeNode(int val, TreeNode left, TreeNode right) {
91+
* this.val = val;
92+
* this.left = left;
93+
* this.right = right;
94+
* }
95+
* }
96+
*/
97+
class Solution {
98+
public int sumRootToLeaf(TreeNode root) {
99+
return dfs(root, 0);
100+
}
101+
102+
private int dfs(TreeNode root, int t) {
103+
if (root == null) {
104+
return 0;
105+
}
106+
t = (t << 1) | root.val;
107+
if (root.left == null && root.right == null) {
108+
return t;
109+
}
110+
return dfs(root.left, t) + dfs(root.right, t);
111+
}
112+
}
113+
```
114+
115+
### **C++**
116+
117+
```cpp
118+
/**
119+
* Definition for a binary tree node.
120+
* struct TreeNode {
121+
* int val;
122+
* TreeNode *left;
123+
* TreeNode *right;
124+
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
125+
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
126+
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
127+
* };
128+
*/
129+
class Solution {
130+
public:
131+
int sumRootToLeaf(TreeNode* root) {
132+
return dfs(root, 0);
133+
}
134+
135+
int dfs(TreeNode* root, int t) {
136+
if (!root) return 0;
137+
t = (t << 1) | root->val;
138+
if (!root->left && !root->right) return t;
139+
return dfs(root->left, t) + dfs(root->right, t);
140+
}
141+
};
142+
```
65143

144+
### **Go**
145+
146+
```go
147+
/**
148+
* Definition for a binary tree node.
149+
* type TreeNode struct {
150+
* Val int
151+
* Left *TreeNode
152+
* Right *TreeNode
153+
* }
154+
*/
155+
func sumRootToLeaf(root *TreeNode) int {
156+
var dfs func(root *TreeNode, t int) int
157+
dfs = func(root *TreeNode, t int) int {
158+
if root == nil {
159+
return 0
160+
}
161+
t = (t << 1) | root.Val
162+
if root.Left == nil && root.Right == nil {
163+
return t
164+
}
165+
return dfs(root.Left, t) + dfs(root.Right, t)
166+
}
167+
168+
return dfs(root, 0)
169+
}
66170
```
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68172
### **...**

‎solution/1000-1099/1022.Sum of Root To Leaf Binary Numbers/Solution.cpp

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/**
2+
* 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 sumRootToLeaf(TreeNode* root) {
15+
return dfs(root, 0);
16+
}
17+
18+
int dfs(TreeNode* root, int t) {
19+
if (!root) return 0;
20+
t = (t << 1) | root->val;
21+
if (!root->left && !root->right) return t;
22+
return dfs(root->left, t) + dfs(root->right, t);
23+
}
24+
};

‎solution/1000-1099/1022.Sum of Root To Leaf Binary Numbers/Solution.go

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1+
/**
2+
* Definition for a binary tree node.
3+
* type TreeNode struct {
4+
* Val int
5+
* Left *TreeNode
6+
* Right *TreeNode
7+
* }
8+
*/
9+
func sumRootToLeaf(root *TreeNode) int {
10+
var dfs func(root *TreeNode, t int) int
11+
dfs = func(root *TreeNode, t int) int {
12+
if root == nil {
13+
return 0
14+
}
15+
t = (t << 1) | root.Val
16+
if root.Left == nil && root.Right == nil {
17+
return t
18+
}
19+
return dfs(root.Left, t) + dfs(root.Right, t)
20+
}
21+
22+
return dfs(root, 0)
23+
}

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