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.
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 pathsum(TreeNode *root,int sum,int val) 13 { 14 if(root==NULL)return 0; 15 val+=root->val; 16 if(root->left==NULL&&root->right==NULL) 17 { 18 if(sum==val)return 1; 19 return 0; 20 21 } 22 23 return pathsum(root->left,sum,val)||pathsum(root->right,sum,val); 24 25 26 } 27 28 29 bool hasPathSum(TreeNode *root, int sum) { 30 31 return pathsum(root,sum,0); 32 } 33 };
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]
]
1 class Solution { 2 private: 3 vector<vector<int> > ret; 4 public: 5 void dfs(TreeNode *node,int sum,int curSum,vector<int> tmp) 6 { 7 if(node==NULL) return; 8 9 if(node->left==NULL &&node->right==NULL) 10 { 11 if(curSum+node->val==sum) 12 { 13 tmp.push_back(node->val); 14 ret.push_back(tmp); 15 } 16 return ; 17 } 18 19 tmp.push_back(node->val); 20 dfs(node->left,sum,curSum+node->val,tmp); 21 dfs(node->right,sum,curSum+node->val,tmp); 22 } 23 24 vector<vector<int>> pathSum(TreeNode* root, int sum) { 25 ret.clear(); 26 vector<int> tmp; 27 dfs(root,sum,0,tmp); 28 return ret; 29 } 30 };
