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Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:sum = 22,
5
/ 4 8
/ / 11 13 4
/ \ 7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
/** * 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: bool hasPathSum(TreeNode* root,int sum,int curSum){ if(!root) return false; curSum += root->val; if(root->left==NULL && root->right==NULL){ cout<<curSum<<endl; if(curSum == sum) return true; return false; } return hasPathSum(root->left,sum,curSum) || hasPathSum(root->right,sum,curSum); } bool hasPathSum(TreeNode* root, int sum) { return hasPathSum(root,sum,0); } };
Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum.
For example:sum = 22,
5
/ 4 8
/ / 11 13 4
/ \ / 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
/** * 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 { private: void pathSum(TreeNode* root, vector<vector<int>>& res, vector<int>& oneRes,int sum,int curSum) { if(!root) return; curSum += root->val; oneRes.push_back(root->val); if(!root->left && !root->right){ if(curSum == sum){ res.push_back(oneRes); } oneRes.pop_back(); return; } pathSum(root->left,res,oneRes,sum,curSum); pathSum(root->right,res,oneRes,sum,curSum); oneRes.pop_back(); } public: vector<vector<int>> pathSum(TreeNode* root, int sum) { vector<vector<int>> res; vector<int> oneRes; pathSum(root,res,oneRes,sum,0); return res; } };
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原文地址：http://www.cnblogs.com/zengzy/p/5053494.html
