題目描述(Hard)

Given a non-empty binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.

題目連結

Example 1:

Input: [1,2,3]

       1       / \      2   3

Output: 6

Example 2:

Input: [-10,9,20,null,null,15,7]

   -10    / \   9  20     /  \    15   7

Output: 42

演算法分析

類似【劍指】42.連續子陣列的最大和，對於有兩個方向的二叉樹，先算出左右子樹的結果L和R，如果L>0，那麼對於後續結果是有利的，我們加上L，R同理。最後返回值應只返回一個方向上的值，因為遞迴只能向父節點返回。

如Example2中所示，路徑為15->20->7。

提交程式碼：

class Solution {
public:
    int maxPathSum(TreeNode* root) {
        int max_sum = INT_MIN;
        dfs(root, max_sum);
        return max_sum;
    }
    
    int dfs(TreeNode* root, int& max_sum) {
        if (!root) return 0;
        
        int left = dfs(root->left, max_sum);
        int right = dfs(root->right, max_sum);
        
        int curr = root->val;
        if (left > 0) curr += left;
        if (right > 0) curr += right;
        
        max_sum = max(max_sum, curr);
        
        return max(left, right) > 0 ? max(left, right) + root->val : root->val;
    }
};