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
题解：
此题类似于字符串或者数组的区间问题，区间和、乘积等。最直接的思想就是枚举所有可能区间。此题先遍历树，在遍历过程中对遍历节点枚举以其为始点的所有区间的和是否等于sum。
Solution 1
1 class Solution { 2 public: 3 int pathSum(TreeNode* root, int sum) { 4 if (!root) 5 return 0; 6 int res = 0; 7 stack<TreeNode*> st; 8 TreeNode* cur = root; 9 while (cur || !st.empty()) { 10 if (cur) { 11 st.push(cur); 12 cur = cur->left; 13 } else { 14 cur = st.top(); 15 st.pop(); 16 int num = 0; 17 dfs(cur, num, sum); 18 res += num; 19 cur = cur->right; 20 } 21 } 22 return res; 23 } 24 void dfs(TreeNode* root, int& num, int sum) { 25 if (!root) 26 return; 27 if (root->val == sum) { 28 ++num; 29 } 30 dfs(root->left, num, sum - root->val); 31 dfs(root->right, num, sum - root->val); 32 33 } 34 };
Solution 2
以root为根节点的树的sum = 以root为起点的区间满足条件的个数 + 以root左孩子为根节点的树的sum + 以root右孩子根节点的树的sum，得到递推关系。
注意 dfs 搜索的是以root为起点的区间，而pathSum搜索的是以root为根节点的树，pathSum没有要求区间必须以root为起点。
1 class Solution { 2 public: 3 int pathSum(TreeNode* root, int sum) { 4 if (!root) 5 return 0; 6 return dfs(root, sum) + pathSum(root->left, sum) + pathSum(root->right, sum); 7 } 8 int dfs(TreeNode* root, int sum) { 9 int num = 0; 10 if (!root) 11 return 0; 12 if (root->val == sum) { 13 ++num; 14 } 15 num += dfs(root->left, sum - root->val); 16 num += dfs(root->right, sum - root->val); 17 return num; 18 } 19 };
Solution 3
思路用的是数组字符串区间常用的哈希存储思想，一般来讲树上不太常用，严格来讲应该和前缀和有些相似。Grandyang
1 class Solution { 2 public: 3 int pathSum(TreeNode* root, int sum) { 4 unordered_map<int, int> m; 5 m[0] = 1; 6 return helper(root, sum, 0, m); 7 } 8 9 int helper(TreeNode* node, int sum, int curSum, unordered_map<int, int>& m) { 10 if (!node) 11 return 0; 12 curSum += node->val; 13 int res = m[curSum - sum]; 14 ++m[curSum]; 15 res += helper(node->left, sum, curSum, m) + helper(node->right, sum, curSum, m); 16 --m[curSum]; 17 return res; 18 } 19 };