Q:
You are given a binary tree in which each node contains an integer value.
Find the number of paths that sum to a given value.
The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).
The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.
Example:
root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8 10 / \ 5 -3 / \ \ 3 2 11 / \ \ 3 -2 1 Return 3. The paths that sum to 8 are: 1. 5 -> 3 2. 5 -> 2 -> 1 3. -3 -> 11
A:
这个是错的,具体是为什么呢? 答案看最下方/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int pathSum(TreeNode* root, int sum) { if(!root) return 0; int endHere = sum == root->val? 1:0; return pathSum(root->left, sum-root->val) +
pathSum(root->right, sum-root->val) +
pathSum(root->left, sum) +
pathSum(root->right, sum) + endHere; } };
my real solution:
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int pathSum(TreeNode* root, int sum) { if(!root) return 0; return pathSum(root->left, sum) +
pathSum(root->right, sum) +
directSum(root,sum); } private: int directSum(TreeNode* root, int sum) { if(!root) return 0; int endHere = sum==root->val?1:0; return directSum(root->left, sum-root->val) + directSum(root->right, sum-root->val) + endHere; } };
为啥错呢? 因为在递归的时候,把一些路径计算了2次。
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