Binary Tree Maximum Path Sum
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.
解析
构建要给helper辅助函数,在递归函数中,如果当前结点不存在,那么直接返回0。否则就分别对其左右子节点调用递归函数,由于路径和有可能为负数,而我们当然不希望加上负的路径和,所以我们和0相比,取较大的那个,就是要么不加,加就要加正数。然后我们来更新全局最大值结果res,就是以左子结点为终点的最大path之和加上以右子结点为终点的最大path之和,还要加上当前结点值,这样就组成了一个条完整的路径。而我们返回值是取left和right中的较大值加上当前结点值,因为我们返回值的定义是以当前结点为终点的path之和,所以只能取left和right中较大的那个值,而不是两个都要。
代码
class Solution {
public:
int maxPathSum(TreeNode* root) {
int res = INT_MIN;
helper(root, res);
return res;
}
int helper(TreeNode* root, int& res){
if(!root) return 0;
int left = max(helper(root->left, res), 0);
int right = max(helper(root->right, res), 0);
res = max(res, left + right + root->val);
return max(left, right) + root->val;
}
};