Path Sum I
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.
递归求解就可以。注意叶子结点条件是左右结点都为空。
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 if (root == NULL) return false; 14 return recursive(root, sum, 0); 15 } 16 17 bool recursive(TreeNode *root, int sum, int cur) { 18 if (root == NULL) { 19 return false; 20 } 21 22 cur += root->val; 23 24 // leaf 25 if (root->left == NULL && root->right == NULL) { 26 return cur == sum; 27 } 28 29 bool res = recursive(root->left, sum, cur); 30 if (res) return true; 31 return recursive(root->right, sum, cur); 32 } 33 };
第三次刷写得简洁许多，我今天已经重复了好多次这一句了。。。
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 if (root == NULL) return false; 14 sum -= root->val; 15 if (root->left == NULL && root->right == NULL && sum == 0) return true; 16 return (hasPathSum(root->left, sum) || hasPathSum(root->right, sum)); 17 } 18 };
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] ]
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 if (root == NULL) return ret; 14 vector<int> v; 15 16 int curSum = 0; 17 recursive(root, sum, curSum, v); 18 return ret; 19 } 20 21 void recursive(TreeNode *root, int sum, int curSum, vector<int> &v) { 22 if (root == NULL) return; 23 24 curSum += root->val; 25 v.push_back(root->val); 26 if (root->left == NULL && root->right == NULL) { 27 if (curSum == sum) { 28 ret.push_back(v); 29 } 30 } 31 recursive(root->left, sum, curSum, v); 32 recursive(root->right, sum, curSum, v); 33 v.pop_back(); 34 } 35 36 private: 37 vector<vector<int> > ret; 38 };
第三次。
1 class Solution { 2 public: 3 void recurse(TreeNode *root, int sum, vector<int> &temp, vector<vector<int> > &ans) { 4 if (root == NULL) return; 5 sum -= root->val; 6 temp.push_back(root->val); 7 if (root->left == NULL && root->right == NULL && sum == 0) { 8 ans.push_back(temp); 9 } else { 10 recurse(root->left, sum, temp, ans); 11 recurse(root->right, sum, temp, ans); 12 } 13 temp.pop_back(); 14 } 15 vector<vector<int> > pathSum(TreeNode *root, int sum) { 16 vector<vector<int> > ans; 17 vector<int> temp; 18 recurse(root, sum, temp, ans); 19 return ans; 20 } 21 };