題目

分析

和上一篇LeetCode算法題很像。使用遞歸來解。注意只有一個節點時且節點為負數，不能直接返回0
Time Complexity:O(n)
Space ComplexityO(h)

流程

定義一個子函數分別處理子樹返回子樹的sum，記錄最大和。 

代碼

/**
 * 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 maxPathSum(TreeNode* root) {
        int ans = INT_MIN;
        maxPathFun(root, ans);
        return ans;
    }
    int maxPathFun(TreeNode* root, int& ans){
        if(!root)
            return 0;
        int l = max(0, maxPathFun(root->left, ans));
        int r = max(0, maxPathFun(root->right, ans));
        int sum = root->val + l + r;
        ans = max(ans, sum);
        return root->val + max(l, r);
        
    }
};

python版

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def __init__(self):
        self.ans = float('-inf')
    def maxPathSum(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        def sum(root):
            if root is None:
                return 0
            l = max(0, sum(root.left))
            r = max(0, sum(root.right))
            s = l + r + root.val
            self.ans = max(self.ans, s)
            return max(l, r) + root.val
        sum(root)
        return self.ans