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:
Given the below binary tree and 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.
递归求解，代码如下：
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool hasPathSum(TreeNode *root, int sum) { 13 // IMPORTANT: Please reset any member data you declared, as 14 // the same Solution instance will be reused for each test case. 15 if(root == NULL)return false; 16 if(root->left == NULL && root->right == NULL && root->val == sum) 17 return true; 18 if(root->left && hasPathSum(root->left, sum - root->val)) 19 return true; 20 if(root->right && hasPathSum(root->right, sum - root->val)) 21 return true; 22 return false; 23 } 24 };
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]
] 本文地址
同理也是递归求解，只是要保存当前的结果，并且每次递归出来后要恢复递归前的结果，每当递归到叶子节点时就把当前结果保存下来。
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int> > pathSum(TreeNode *root, int sum) { 13 // IMPORTANT: Please reset any member data you declared, as 14 // the same Solution instance will be reused for each test case. 15 vector<vector<int> > res; 16 if(root == NULL)return res; 17 vector<int> curres; 18 curres.push_back(root->val); 19 pathSumRecur(root, sum, curres, res); 20 return res; 21 } 22 void pathSumRecur(TreeNode *root, int sum, vector<int> &curres, vector<vector<int> >&res) 23 { 24 if(root->left == NULL && root->right == NULL && root->val == sum) 25 { 26 res.push_back(curres); 27 return; 28 } 29 if(root->left) 30 { 31 curres.push_back(root->left->val); 32 pathSumRecur(root->left, sum - root->val, curres, res); 33 curres.pop_back(); 34 } 35 if(root->right) 36 { 37 curres.push_back(root->right->val); 38 pathSumRecur(root->right, sum - root->val, curres, res); 39 curres.pop_back(); 40 } 41 } 42 };
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