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feat: add solutions to lc problem: No.0298
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No.0298.Binary Tree Longest Consecutive Sequence
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yanglbme committed Dec 24, 2021
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180 changes: 149 additions & 31 deletions solution/0200-0299/0298.Binary Tree Longest Consecutive Sequence/README.md
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<!-- 这里写题目描述 -->

<p>给你一棵指定的二叉树，请你计算它最长连续序列路径的长度。</p>
<p>给你一棵指定的二叉树的根节点 <code>root</code> ，请你计算其中 <strong>最长连续序列路径</strong> 的长度。</p>

<p>该路径，可以是从某个初始结点到树中任意结点，通过「父 - 子」关系连接而产生的任意路径。</p>

<p>这个最长连续的路径，必须从父结点到子结点，反过来是不可以的。</p>
<p><strong>最长连续序列路径</strong> 是依次递增 1 的路径。该路径，可以是从某个初始节点到树中任意节点，通过「父 - 子」关系连接而产生的任意路径。且必须从父节点到子节点，反过来是不可以的。</p>
&nbsp;

<p><strong>示例 1：</strong></p>

<pre><strong>输入:</strong>

1
\
3
/ \
2 4
\
5

<strong>输出:</strong> <code>3</code>

<strong>解析: </strong>当中，最长连续序列是 <code>3-4-5，所以</code>返回结果为 <code>3</code></pre>
<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0298.Binary%20Tree%20Longest%20Consecutive%20Sequence/images/consec1-1-tree.jpg" style="width: 322px; height: 421px;" />
<pre>
<strong>输入：</strong>root = [1,null,3,2,4,null,null,null,5]
<strong>输出：</strong>3
<strong>解释：</strong>当中，最长连续序列是 <code>3-4-5 ，所以</code>返回结果为 <code>3 。</code>
</pre>

<p><strong>示例 2：</strong></p>
<img alt="" src="https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0200-0299/0298.Binary%20Tree%20Longest%20Consecutive%20Sequence/images/consec1-2-tree.jpg" style="width: 262px; height: 421px;" />
<pre>
<strong>输入：</strong>root = [2,null,3,2,null,1]
<strong>输出：</strong>2
<strong>解释：</strong>当中，最长连续序列是 <code>2-3 。注意，不是</code> <code>3-2-1，所以</code>返回 <code>2 。</code>
</pre>

<pre><strong>输入:

</strong> 2
\
3
/
2
/
1
<p>&nbsp;</p>

<strong>输出: 2

解析: </strong>当中，最长连续序列是 <code>2-3。注意，不是</code> <code>3-2-1，所以</code>返回 <code>2。</code></pre>
<p><strong>提示：</strong></p>

<ul>
<li>树中节点的数目在范围 <code>[1, 3 * 10<sup>4</sup>]</code> 内</li>
<li><code>-3 * 10<sup>4</sup> &lt;= Node.val &lt;= 3 * 10<sup>4</sup></code></li>
</ul>

## 解法

<!-- 这里可写通用的实现逻辑 -->

DFS。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```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 longestConsecutive(self, root: TreeNode) -> int:
def dfs(root, p, t):
nonlocal ans
if root is None:
return
t = t + 1 if p is not None and p.val + 1 == root.val else 1
ans = max(ans, t)
dfs(root.left, root, t)
dfs(root.right, root, t)

ans = 1
dfs(root, None, 1)
return ans
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private int ans;

public int longestConsecutive(TreeNode root) {
ans = 1;
dfs(root, null, 1);
return ans;
}

private void dfs(TreeNode root, TreeNode p, int t) {
if (root == null) {
return;
}
t = p != null && p.val + 1 == root.val ? t + 1 : 1;
ans = Math.max(ans, t);
dfs(root.left, root, t);
dfs(root.right, root, t);
}
}
```

### **C++**

```cpp
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int ans;

int longestConsecutive(TreeNode* root) {
ans = 1;
dfs(root, nullptr, 1);
return ans;
}

void dfs(TreeNode* root, TreeNode* p, int t) {
if (!root) return;
t = p != nullptr && p->val + 1 == root-> val ? t + 1 : 1;
ans = max(ans, t);
dfs(root->left, root, t);
dfs(root->right, root, t);
}
};
```
### **Go**
```go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func longestConsecutive(root *TreeNode) int {
ans := 1
var dfs func(root, p *TreeNode, t int)
dfs = func(root, p *TreeNode, t int) {
if root == nil {
return
}
if p != nil && p.Val+1 == root.Val {
t++
ans = max(ans, t)
} else {
t = 1
}
dfs(root.Left, root, t)
dfs(root.Right, root, t)
}
dfs(root, nil, 1)
return ans
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
```

### **...**
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130 changes: 128 additions & 2 deletions solution/0200-0299/0298.Binary Tree Longest Consecutive Sequence/README_EN.md
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<li><code>-3 * 10<sup>4</sup> &lt;= Node.val &lt;= 3 * 10<sup>4</sup></code></li>
</ul>


## Solutions

DFS.

<!-- 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 longestConsecutive(self, root: TreeNode) -> int:
def dfs(root, p, t):
nonlocal ans
if root is None:
return
t = t + 1 if p is not None and p.val + 1 == root.val else 1
ans = max(ans, t)
dfs(root.left, root, t)
dfs(root.right, root, t)

ans = 1
dfs(root, None, 1)
return ans
```

### **Java**

```java
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private int ans;

public int longestConsecutive(TreeNode root) {
ans = 1;
dfs(root, null, 1);
return ans;
}

private void dfs(TreeNode root, TreeNode p, int t) {
if (root == null) {
return;
}
t = p != null && p.val + 1 == root.val ? t + 1 : 1;
ans = Math.max(ans, t);
dfs(root.left, root, t);
dfs(root.right, root, t);
}
}
```

### **C++**

```cpp
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int ans;

int longestConsecutive(TreeNode* root) {
ans = 1;
dfs(root, nullptr, 1);
return ans;
}

void dfs(TreeNode* root, TreeNode* p, int t) {
if (!root) return;
t = p != nullptr && p->val + 1 == root-> val ? t + 1 : 1;
ans = max(ans, t);
dfs(root->left, root, t);
dfs(root->right, root, t);
}
};
```
### **Go**
```go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func longestConsecutive(root *TreeNode) int {
ans := 1
var dfs func(root, p *TreeNode, t int)
dfs = func(root, p *TreeNode, t int) {
if root == nil {
return
}
if p != nil && p.Val+1 == root.Val {
t++
ans = max(ans, t)
} else {
t = 1
}
dfs(root.Left, root, t)
dfs(root.Right, root, t)
}
dfs(root, nil, 1)
return ans
}
func max(a, b int) int {
if a > b {
return a
}
return b
}
```

### **...**
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29 changes: 29 additions & 0 deletions solution/0200-0299/0298.Binary Tree Longest Consecutive Sequence/Solution.cpp
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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int ans;

int longestConsecutive(TreeNode* root) {
ans = 1;
dfs(root, nullptr, 1);
return ans;
}

void dfs(TreeNode* root, TreeNode* p, int t) {
if (!root) return;
t = p != nullptr && p->val + 1 == root-> val ? t + 1 : 1;
ans = max(ans, t);
dfs(root->left, root, t);
dfs(root->right, root, t);
}
};
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