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feat: add solutions to lc/lcof2 problem: Binary Tree Pruning
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‎lcof2/剑指 Offer II 047. 二叉树剪枝/README.md

+70-5
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@@ -14,8 +14,8 @@
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<pre>
1616
<strong>输入:</strong> [1,null,0,0,1]
17-
<strong>输出: </strong>[1,null,0,null,1]
18-
<strong>解释:</strong>
17+
<strong>输出: </strong>[1,null,0,null,1]
18+
<strong>解释:</strong>
1919
只有红色节点满足条件&ldquo;所有不包含 1 的子树&rdquo;。
2020
右图为返回的答案。
2121

@@ -27,7 +27,7 @@
2727
<pre>
2828
<strong>输入:</strong> [1,0,1,0,0,0,1]
2929
<strong>输出: </strong>[1,null,1,null,1]
30-
<strong>解释:</strong>
30+
<strong>解释:</strong>
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<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/lcof2/%E5%89%91%E6%8C%87%20Offer%20II%20047.%20%E4%BA%8C%E5%8F%89%E6%A0%91%E5%89%AA%E6%9E%9D/images/1028_1.png" style="width:450px" />
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</pre>
@@ -37,7 +37,7 @@
3737
<pre>
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<strong>输入:</strong> [1,1,0,1,1,0,1,0]
3939
<strong>输出: </strong>[1,1,0,1,1,null,1]
40-
<strong>解释:</strong>
40+
<strong>解释:</strong>
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4242
<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/lcof2/%E5%89%91%E6%8C%87%20Offer%20II%20047.%20%E4%BA%8C%E5%8F%89%E6%A0%91%E5%89%AA%E6%9E%9D/images/1028.png" style="width:450px" />
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</pre>
@@ -67,15 +67,80 @@
6767
<!-- 这里可写当前语言的特殊实现逻辑 -->
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6969
```python
70-
70+
# Definition for a binary tree node.
71+
# class TreeNode:
72+
# def __init__(self, val=0, left=None, right=None):
73+
# self.val = val
74+
# self.left = left
75+
# self.right = right
76+
class Solution:
77+
def pruneTree(self, root: TreeNode) -> TreeNode:
78+
if not root:
79+
return None
80+
root.left = self.pruneTree(root.left)
81+
root.right = self.pruneTree(root.right)
82+
if root.val == 0 and not root.left and not root.right:
83+
return None
84+
return root
7185
```
7286

7387
### **Java**
7488

7589
<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
92+
/**
93+
* Definition for a binary tree node.
94+
* public class TreeNode {
95+
* int val;
96+
* TreeNode left;
97+
* TreeNode right;
98+
* TreeNode() {}
99+
* TreeNode(int val) { this.val = val; }
100+
* TreeNode(int val, TreeNode left, TreeNode right) {
101+
* this.val = val;
102+
* this.left = left;
103+
* this.right = right;
104+
* }
105+
* }
106+
*/
107+
class Solution {
108+
public TreeNode pruneTree(TreeNode root) {
109+
if (root == null) {
110+
return null;
111+
}
112+
root.left = pruneTree(root.left);
113+
root.right = pruneTree(root.right);
114+
if (root.val == 0 && root.left == null && root.right == null) {
115+
return null;
116+
}
117+
return root;
118+
}
119+
}
120+
```
78121

122+
### **Go**
123+
124+
```go
125+
/**
126+
* Definition for a binary tree node.
127+
* type TreeNode struct {
128+
* Val int
129+
* Left *TreeNode
130+
* Right *TreeNode
131+
* }
132+
*/
133+
func pruneTree(root *TreeNode) *TreeNode {
134+
if root == nil {
135+
return nil
136+
}
137+
root.Left = pruneTree(root.Left)
138+
root.Right = pruneTree(root.Right)
139+
if root.Val == 0 && root.Left == nil && root.Right == nil {
140+
return nil
141+
}
142+
return root
143+
}
79144
```
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81146
### **...**

‎lcof2/剑指 Offer II 047. 二叉树剪枝/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 pruneTree(root *TreeNode) *TreeNode {
10+
if root == nil {
11+
return nil
12+
}
13+
root.Left = pruneTree(root.Left)
14+
root.Right = pruneTree(root.Right)
15+
if root.Val == 0 && root.Left == nil && root.Right == nil {
16+
return nil
17+
}
18+
return root
19+
}

‎lcof2/剑指 Offer II 047. 二叉树剪枝/Solution.java

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1+
/**
2+
* Definition for a binary tree node.
3+
* public class TreeNode {
4+
* 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 TreeNode pruneTree(TreeNode root) {
18+
if (root == null) {
19+
return null;
20+
}
21+
root.left = pruneTree(root.left);
22+
root.right = pruneTree(root.right);
23+
if (root.val == 0 && root.left == null && root.right == null) {
24+
return null;
25+
}
26+
return root;
27+
}
28+
}

‎lcof2/剑指 Offer II 047. 二叉树剪枝/Solution.py

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@@ -0,0 +1,15 @@
1+
# Definition for a binary tree node.
2+
# class TreeNode:
3+
# 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 pruneTree(self, root: TreeNode) -> TreeNode:
9+
if not root:
10+
return None
11+
root.left = self.pruneTree(root.left)
12+
root.right = self.pruneTree(root.right)
13+
if root.val == 0 and not root.left and not root.right:
14+
return None
15+
return root

‎solution/0800-0899/0814.Binary Tree Pruning/README.md

+68-3
Original file line numberDiff line numberDiff line change
@@ -16,8 +16,8 @@
1616
<strong>示例1:</strong>
1717
<strong>输入:</strong> [1,null,0,0,1]
1818
<strong>输出: </strong>[1,null,0,null,1]
19-
20-
<strong>解释:</strong>
19+
20+
<strong>解释:</strong>
2121
只有红色节点满足条件&ldquo;所有不包含 1 的子树&rdquo;。
2222
右图为返回的答案。
2323

@@ -61,15 +61,80 @@
6161
<!-- 这里可写当前语言的特殊实现逻辑 -->
6262

6363
```python
64-
64+
# Definition for a binary tree node.
65+
# class TreeNode:
66+
# def __init__(self, val=0, left=None, right=None):
67+
# self.val = val
68+
# self.left = left
69+
# self.right = right
70+
class Solution:
71+
def pruneTree(self, root: TreeNode) -> TreeNode:
72+
if not root:
73+
return None
74+
root.left = self.pruneTree(root.left)
75+
root.right = self.pruneTree(root.right)
76+
if root.val == 0 and not root.left and not root.right:
77+
return None
78+
return root
6579
```
6680

6781
### **Java**
6882

6983
<!-- 这里可写当前语言的特殊实现逻辑 -->
7084

7185
```java
86+
/**
87+
* Definition for a binary tree node.
88+
* public class TreeNode {
89+
* int val;
90+
* TreeNode left;
91+
* TreeNode right;
92+
* TreeNode() {}
93+
* TreeNode(int val) { this.val = val; }
94+
* TreeNode(int val, TreeNode left, TreeNode right) {
95+
* this.val = val;
96+
* this.left = left;
97+
* this.right = right;
98+
* }
99+
* }
100+
*/
101+
class Solution {
102+
public TreeNode pruneTree(TreeNode root) {
103+
if (root == null) {
104+
return null;
105+
}
106+
root.left = pruneTree(root.left);
107+
root.right = pruneTree(root.right);
108+
if (root.val == 0 && root.left == null && root.right == null) {
109+
return null;
110+
}
111+
return root;
112+
}
113+
}
114+
```
72115

116+
### **Go**
117+
118+
```go
119+
/**
120+
* Definition for a binary tree node.
121+
* type TreeNode struct {
122+
* Val int
123+
* Left *TreeNode
124+
* Right *TreeNode
125+
* }
126+
*/
127+
func pruneTree(root *TreeNode) *TreeNode {
128+
if root == nil {
129+
return nil
130+
}
131+
root.Left = pruneTree(root.Left)
132+
root.Right = pruneTree(root.Right)
133+
if root.Val == 0 && root.Left == nil && root.Right == nil {
134+
return nil
135+
}
136+
return root
137+
}
73138
```
74139

75140
### **...**

‎solution/0800-0899/0814.Binary Tree Pruning/README_EN.md

+68-3
Original file line numberDiff line numberDiff line change
@@ -14,8 +14,8 @@
1414
<strong>Example 1:</strong>
1515
<strong>Input:</strong> [1,null,0,0,1]
1616
<strong>Output: </strong>[1,null,0,null,1]
17-
18-
<strong>Explanation:</strong>
17+
18+
<strong>Explanation:</strong>
1919
Only the red nodes satisfy the property &quot;every subtree not containing a 1&quot;.
2020
The diagram on the right represents the answer.
2121

@@ -55,13 +55,78 @@ The diagram on the right represents the answer.
5555
### **Python3**
5656

5757
```python
58-
58+
# Definition for a binary tree node.
59+
# class TreeNode:
60+
# def __init__(self, val=0, left=None, right=None):
61+
# self.val = val
62+
# self.left = left
63+
# self.right = right
64+
class Solution:
65+
def pruneTree(self, root: TreeNode) -> TreeNode:
66+
if not root:
67+
return None
68+
root.left = self.pruneTree(root.left)
69+
root.right = self.pruneTree(root.right)
70+
if root.val == 0 and not root.left and not root.right:
71+
return None
72+
return root
5973
```
6074

6175
### **Java**
6276

6377
```java
78+
/**
79+
* Definition for a binary tree node.
80+
* public class TreeNode {
81+
* int val;
82+
* TreeNode left;
83+
* TreeNode right;
84+
* TreeNode() {}
85+
* TreeNode(int val) { this.val = val; }
86+
* TreeNode(int val, TreeNode left, TreeNode right) {
87+
* this.val = val;
88+
* this.left = left;
89+
* this.right = right;
90+
* }
91+
* }
92+
*/
93+
class Solution {
94+
public TreeNode pruneTree(TreeNode root) {
95+
if (root == null) {
96+
return null;
97+
}
98+
root.left = pruneTree(root.left);
99+
root.right = pruneTree(root.right);
100+
if (root.val == 0 && root.left == null && root.right == null) {
101+
return null;
102+
}
103+
return root;
104+
}
105+
}
106+
```
64107

108+
### **Go**
109+
110+
```go
111+
/**
112+
* Definition for a binary tree node.
113+
* type TreeNode struct {
114+
* Val int
115+
* Left *TreeNode
116+
* Right *TreeNode
117+
* }
118+
*/
119+
func pruneTree(root *TreeNode) *TreeNode {
120+
if root == nil {
121+
return nil
122+
}
123+
root.Left = pruneTree(root.Left)
124+
root.Right = pruneTree(root.Right)
125+
if root.Val == 0 && root.Left == nil && root.Right == nil {
126+
return nil
127+
}
128+
return root
129+
}
65130
```
66131

67132
### **...**

‎solution/0800-0899/0814.Binary Tree Pruning/Solution.go

+19
Original file line numberDiff line numberDiff line change
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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 pruneTree(root *TreeNode) *TreeNode {
10+
if root == nil {
11+
return nil
12+
}
13+
root.Left = pruneTree(root.Left)
14+
root.Right = pruneTree(root.Right)
15+
if root.Val == 0 && root.Left == nil && root.Right == nil {
16+
return nil
17+
}
18+
return root
19+
}

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