437. Path Sum III
2017-01-09 22:43
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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:
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以每一个节点作为路径根节点进行前序遍历,查找每一条路径的权值和与sum是否相等
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
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以每一个节点作为路径根节点进行前序遍历,查找每一条路径的权值和与sum是否相等
public: int pathSum(TreeNode* root, int sum) { if(!root) return 0; return sumUp(root, 0, sum) + pathSum(root->left, sum) + pathSum(root->right, sum); } private: int sumUp(TreeNode* root, int pre, int& sum){ if(!root) return 0; int current = pre + root->val; return (current == sum) + sumUp(root->left, current, sum) + sumUp(root->right, current, sum); }详细多种方法见:http://www.cnblogs.com/VickyWang/p/6038386.html
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