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feat: add solutions to lc/lcof2 problem: Binary Tree Maximum Path Sum
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‎lcof2/剑指 Offer II 051. 节点之和最大的路径/README.md

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<!-- 这里可写通用的实现逻辑 -->
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思考二叉树递归问题的经典套路：
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1. 终止条件（何时终止递归）
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2. 递归处理左右子树
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3. 合并左右子树的计算结果
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对于本题，由于要满足题目对“路径”的定义，在返回当前子树对外贡献的最大路径和时，需要取 `left`，`right` 的最大值
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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 maxPathSum(self, root: TreeNode) -> int:
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ans = float("-inf")
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def dfs(node: TreeNode) -> int:
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if not node:
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return 0
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left = max(0, dfs(node.left))
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right = max(0, dfs(node.right))
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nonlocal ans
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ans = max(ans, node.val + left + right)
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return node.val + max(left, right)
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dfs(root)
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return ans
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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 ans = Integer.MIN_VALUE;
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public int maxPathSum(TreeNode root) {
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dfs(root);
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return ans;
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}
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private int dfs(TreeNode node) {
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if (node == null) {
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return 0;
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}
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int left = Math.max(0, dfs(node.left));
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int right = Math.max(0, dfs(node.right));
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ans = Math.max(ans, node.val + left + right);
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return node.val + Math.max(left, 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 maxPathSum(root *TreeNode) int {
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ans := math.MinInt32
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var dfs func(*TreeNode) int
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dfs = func(node *TreeNode) int {
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if node == nil {
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return 0
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}
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left := max(0, dfs(node.Left))
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right := max(0, dfs(node.Right))
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ans = max(ans, node.Val+left+right)
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return node.Val + max(left, right)
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}
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dfs(root)
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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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 maxPathSum(TreeNode* root) {
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int ans = INT_MIN;
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function<int(TreeNode*)> dfs = [&](TreeNode* node) {
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if (node == nullptr) {
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return 0;
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}
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int left = max(0, dfs(node->left));
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int right = max(0, dfs(node->right));
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ans = max(ans, node->val + left + right);
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return node->val + max(left, right);
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};
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dfs(root);
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return ans;
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}
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};
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```
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### **...**

‎lcof2/剑指 Offer II 051. 节点之和最大的路径/Solution.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 maxPathSum(TreeNode* root) {
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int ans = INT_MIN;
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function<int(TreeNode*)> dfs = [&](TreeNode* node) {
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if (node == nullptr) {
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return 0;
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}
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int left = max(0, dfs(node->left));
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int right = max(0, dfs(node->right));
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ans = max(ans, node->val + left + right);
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return node->val + max(left, right);
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};
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dfs(root);
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return ans;
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}
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};

‎lcof2/剑指 Offer II 051. 节点之和最大的路径/Solution.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 maxPathSum(root *TreeNode) int {
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ans := math.MinInt32
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var dfs func(*TreeNode) int
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dfs = func(node *TreeNode) int {
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if node == nil {
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return 0
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}
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left := max(0, dfs(node.Left))
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right := max(0, dfs(node.Right))
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ans = max(ans, node.Val+left+right)
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return node.Val + max(left, right)
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}
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dfs(root)
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}

‎lcof2/剑指 Offer II 051. 节点之和最大的路径/Solution.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 ans = Integer.MIN_VALUE;
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public int maxPathSum(TreeNode root) {
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dfs(root);
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return ans;
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}
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private int dfs(TreeNode node) {
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if (node == null) {
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return 0;
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}
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int left = Math.max(0, dfs(node.left));
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int right = Math.max(0, dfs(node.right));
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ans = Math.max(ans, node.val + left + right);
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return node.val + Math.max(left, right);
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}
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}

‎lcof2/剑指 Offer II 051. 节点之和最大的路径/Solution.py

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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 maxPathSum(self, root: TreeNode) -> int:
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ans = float("-inf")
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def dfs(node: TreeNode) -> int:
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if not node:
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return 0
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left = max(0, dfs(node.left))
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right = max(0, dfs(node.right))
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nonlocal ans
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ans = max(ans, node.val + left + right)
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return node.val + max(left, right)
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dfs(root)
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return ans

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