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.
For example:
Given the below binary tree and sum = 22
,
5 / 4 8 / / 11 13 4 / / 7 2 5 1
return [[5,4,11,2],[5,8,4,5]]
分析: dfs求解即可
/** * 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: void find(TreeNode* root, int sum,vector<int>& curPath, vector<vector<int>> & res ){ if(root==nullptr) return; if(root->left){ curPath.push_back(root->left->val); find(root->left, sum-root->val, curPath,res); curPath.pop_back(); } if(root->right){ curPath.push_back(root->right->val); find(root->right, sum-root->val,curPath,res); curPath.pop_back(); } if(root->left==nullptr && root->right==nullptr && sum==root->val) res.push_back(curPath); return; } vector<vector<int>> pathSum(TreeNode* root, int sum) { vector<vector<int>> res; if(root==nullptr) return res; vector<int> curPath; curPath.push_back(root->val); find(root, sum, curPath, res); return res; } };