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feat: add solutions to lc problem: No.0965.Univalued Binary Tree
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Lines changed: 297 additions & 37 deletions

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‎solution/0900-0999/0965.Univalued Binary Tree/README.md‎

Lines changed: 106 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -37,7 +37,6 @@
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<li>每个节点的值都是整数，范围为&nbsp;<code>[0, 99]</code>&nbsp;。</li>
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</ol>
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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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<!-- 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 isUnivalTree(self, root: TreeNode) -> bool:
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def dfs(root):
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if root is None:
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return True
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if root.val != self.val:
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return False
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return dfs(root.left) and dfs(root.right)
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self.val = root.val
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return dfs(root)
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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 int val;
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public boolean isUnivalTree(TreeNode root) {
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val = root.val;
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return dfs(root);
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}
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private boolean dfs(TreeNode root) {
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if (root == null) {
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return true;
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}
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if (root.val != val) {
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return false;
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}
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return dfs(root.left) && dfs(root.right);
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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 val;
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bool isUnivalTree(TreeNode* root) {
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val = root->val;
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return dfs(root);
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}
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bool dfs(TreeNode* root) {
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if (root == nullptr) return true;
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if (root->val != val) return false;
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return dfs(root->left) && dfs(root->right);
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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 isUnivalTree(root *TreeNode) bool {
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return dfs(root, root.Val)
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}
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func dfs(root *TreeNode, val int) bool {
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if root == nil {
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return true
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}
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if root.Val != val {
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return false
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}
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return dfs(root.Left, val) && dfs(root.Right, val)
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}
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```
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### **...**

‎solution/0900-0999/0965.Univalued Binary Tree/README_EN.md‎

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@@ -6,16 +6,10 @@
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<p>A binary tree is <em>univalued</em> if every node in the tree has the same value.</p>
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<p>Return <code>true</code>&nbsp;if and only if the given tree is univalued.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0900-0999/0965.Univalued%20Binary%20Tree/images/unival_bst_1.png" style="width: 265px; height: 172px;" />
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</pre>
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<div>
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<p><strong>Example 2:</strong></p>
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</div>
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<p>&nbsp;</p>
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<p><strong>Note:</strong></p>
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<ol>
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<li>The number of nodes in the given tree will be in the range <code>[1, 100]</code>.</li>
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<li>Each node&#39;s value will be an integer in the range <code>[0, 99]</code>.</li>
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</ol>
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## Solutions
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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:
59+
# def __init__(self, val=0, left=None, right=None):
60+
# self.val = val
61+
# self.left = left
62+
# self.right = right
63+
class Solution:
64+
def isUnivalTree(self, root: TreeNode) -> bool:
65+
def dfs(root):
66+
if root is None:
67+
return True
68+
if root.val != self.val:
69+
return False
70+
return dfs(root.left) and dfs(root.right)
71+
72+
self.val = root.val
73+
return dfs(root)
7474
```
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### **Java**
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```java
79+
/**
80+
* Definition for a binary tree node.
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* public class TreeNode {
82+
* int val;
83+
* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
87+
* TreeNode(int val, TreeNode left, TreeNode right) {
88+
* this.val = val;
89+
* this.left = left;
90+
* this.right = right;
91+
* }
92+
* }
93+
*/
94+
class Solution {
95+
private int val;
96+
97+
public boolean isUnivalTree(TreeNode root) {
98+
val = root.val;
99+
return dfs(root);
100+
}
101+
102+
private boolean dfs(TreeNode root) {
103+
if (root == null) {
104+
return true;
105+
}
106+
if (root.val != val) {
107+
return false;
108+
}
109+
return dfs(root.left) && dfs(root.right);
110+
}
111+
}
112+
```
113+
114+
### **C++**
115+
116+
```cpp
117+
/**
118+
* Definition for a binary tree node.
119+
* struct TreeNode {
120+
* int val;
121+
* TreeNode *left;
122+
* TreeNode *right;
123+
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
124+
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
125+
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
126+
* };
127+
*/
128+
class Solution {
129+
public:
130+
int val;
131+
132+
bool isUnivalTree(TreeNode* root) {
133+
val = root->val;
134+
return dfs(root);
135+
}
136+
137+
bool dfs(TreeNode* root) {
138+
if (root == nullptr) return true;
139+
if (root->val != val) return false;
140+
return dfs(root->left) && dfs(root->right);
141+
}
142+
};
143+
```
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### **Go**
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147+
```go
148+
/**
149+
* Definition for a binary tree node.
150+
* type TreeNode struct {
151+
* Val int
152+
* Left *TreeNode
153+
* Right *TreeNode
154+
* }
155+
*/
156+
func isUnivalTree(root *TreeNode) bool {
157+
return dfs(root, root.Val)
158+
}
159+
160+
func dfs(root *TreeNode, val int) bool {
161+
if root == nil {
162+
return true
163+
}
164+
if root.Val != val {
165+
return false
166+
}
167+
return dfs(root.Left, val) && dfs(root.Right, val)
168+
}
80169
```
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### **...**

‎solution/0900-0999/0965.Univalued Binary Tree/Solution.cpp‎

Lines changed: 14 additions & 18 deletions
Original file line numberDiff line numberDiff line change
@@ -4,27 +4,23 @@
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* int val;
55
* TreeNode *left;
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* TreeNode *right;
7-
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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) {}
810
* };
911
*/
1012
class Solution {
11-
private:
12-
bool isUnivalVal(TreeNode* r, int val)
13-
{
14-
if (val != r->val)
15-
return false ;
16-
17-
if (r->left && !isUnivalVal(r->left, val))
18-
return false ;
19-
if (r->right && !isUnivalVal(r->right, val))
20-
return false ;
21-
22-
return true ;
23-
}
2413
public:
14+
int val;
15+
2516
bool isUnivalTree(TreeNode* root) {
26-
if (nullptr == root)
27-
return true ;
28-
return isUnivalVal(root, root->val) ;
17+
val = root->val;
18+
return dfs(root);
19+
}
20+
21+
bool dfs(TreeNode* root) {
22+
if (root == nullptr) return true;
23+
if (root->val != val) return false;
24+
return dfs(root->left) && dfs(root->right);
2925
}
30-
} ;
26+
};

‎solution/0900-0999/0965.Univalued Binary Tree/Solution.go‎

Lines changed: 21 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,21 @@
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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
7+
* }
8+
*/
9+
func isUnivalTree(root *TreeNode) bool {
10+
return dfs(root, root.Val)
11+
}
12+
13+
func dfs(root *TreeNode, val int) bool {
14+
if root == nil {
15+
return true
16+
}
17+
if root.Val != val {
18+
return false
19+
}
20+
return dfs(root.Left, val) && dfs(root.Right, val)
21+
}

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