LeetCode 124. Binary Tree Maximum Path Sum

Posted 2020-06-17

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int maxPathSum(TreeNode* root) {
13         int partialSum=0;
14         int maxSum=INT_MIN;
15         calculateSum(root, partialSum, maxSum);
16         return maxSum;
17     }
18 
19     //partialSum means the maximal sum which contains only current node and only one of its subtree (either left or right)
20     void calculateSum(TreeNode * root, int & partialSum, int & maxSum) {
21         if(root == NULL) return;
22         int leftPartialSum=0;
23         if(root->left != NULL)
24             calculateSum(root->left, leftPartialSum, maxSum);
25 
26         int rightPartialSum=0;
27         if(root->right != NULL)
28             calculateSum(root->right, rightPartialSum, maxSum);
29 
30         partialSum=max(root->val, max(root->val+leftPartialSum, root->val+rightPartialSum));
31         maxSum=max(partialSum, max(maxSum, leftPartialSum+rightPartialSum+root->val));
32     }
33 };

partialSum仅有如下几种情况：root, root+leftSubtree, root+rightSubtree. 即一条合法的路径。

maxSum有如下几种情况：1、即partialSum 2、左儿子+root+右儿子 3、之前的maxSum值

关于其中第三种情况的解释：maxSum存储了目前为止最大和路径的值,

                                     比如  -1

                                            /    \

                                          -2      3

                                     maxSum先是-2（处理-2节点时），然后是3（处理3时），最后是3（处理-1时）。而处理-1时partialSum为-1,-3,2中最大的，即2.

 

待重写以加深印象，留坑。

注意：                                         

int maxSum = INT_MIN;或int maxSum = 0x80000000

                 

 

另一种写法：

 1 class Solution {
 2 public:
 3 int res;
 4 int maxPathSum(TreeNode* root) {
 5     res = 0x80000000;
 6     dfs(root);
 7     return res;
 8 }
 9 int dfs(TreeNode* tree){
10     if(!tree) return 0;
11     int left = dfs(tree->left), right = dfs(tree->right), tmp;
12     tmp = max(tree->val, tree->val+left);
13     tmp = max(tmp, tree->val+right);
14     res = max(tmp, res);
15     res = max(res, tree->val+left+right);
16     return tmp;
17 
18 }
19 };

 

顺便两份一模一样的代码居然跑的时间不一样，留坑。