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 andsum = 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 // leaf25 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 andsum = 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 > pathSum(TreeNode *root, int sum) {13 if (root == NULL) return ret;14 vector 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 &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 > ret;38 }; 第三次。
1 class Solution { 2 public: 3 void recurse(TreeNode *root, int sum, vector &temp, vector > &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 > pathSum(TreeNode *root, int sum) {16 vector > ans;17 vector temp;18 recurse(root, sum, temp, ans);19 return ans;20 }21 };