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August 29, 2015 14:00
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Path Sum I | |
bool hasPathSum(TreeNode *root, int sum) { | |
return hasPathSumRe(root, 0, sum); | |
} | |
bool hasPathSumRe(TreeNode *root, int value, int sum) { | |
if(!root) return false; | |
if(!root->left && !root->right){ | |
if((value+root->val) == sum) | |
return true; | |
else return false; | |
} | |
else return hasPathSumRe(root->left, root->val + value, sum)||hasPathSumRe(root->right, root->val + value, sum); | |
} | |
Path Sum II | |
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum | |
vector<vector<int> > pathSum(TreeNode *root, int sum) { | |
vector<vector<int>> res; | |
vector<int> onePossibleSol; | |
pathSumRe(root, 0, sum, onePossibleSol, res); | |
return res; | |
} | |
void pathSumRe(TreeNode *root, int value, int sum, vector<int> oneSol, vector<vector<int>> &res) { | |
if(!root) return; | |
if(!root->left && !root->right){ | |
if((value+root->val) == sum){ | |
oneSol.push_back(root->val); | |
res.push_back(oneSol); | |
} | |
else return ; | |
} | |
oneSol.push_back(root->val); | |
pathSumRe(root->left, root->val + value, sum, oneSol, res); | |
pathSumRe(root->right, root->val + value, sum, oneSol, res); | |
} |
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