Leetcode: Binary Tree Maximum Path Sum

思路:

1. 对于每一个节点, 返回其所在子树所能提供的最大值, 且该最大值必须是单支的, WA 过

2. max(0, max(lf, rt))+root->val, 可以仅返回根节点, WA 过

3. 需要维护一个全局最优解 ans, WA 过

代码:

class Solution {
public:
int ans;
int solve_dp(TreeNode *root) {
if(root == NULL)
return 0;

int sum = root->val;
int lf = 0, rt = 0;
if(root->left)
lf = solve_dp(root->left);

if(root->right)
rt = solve_dp(root->right);
if(lf > 0)
sum += lf;
if(rt > 0)
sum += rt;

ans = max(ans, sum);
return max(0, max(lf, rt))+root->val;
}
int maxPathSum(TreeNode *root) {
ans = -100000000;
solve_dp(root);
return ans;
}
};