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] ]
求路径的和与某一特定的值时候对等即可，简单的dfs，代码如下：
1 class Solution { 2 public: 3 vector<vector<int>> pathSum(TreeNode* root, int sum) { 4 vector<int> res; 5 dfs(root, res, sum); 6 return ret; 7 } 8 9 void dfs(TreeNode * root, vector<int> res, int left) 10 { 11 if(!root) return; 12 if(!root->left && !root->right && left == root->val){ 13 res.push_back(root->val); 14 ret.push_back(res); 15 } 16 if(left <= root->val) 17 return; 18 else{ 19 res.push_back(root->val); 20 if(root->left) 21 dfs(root->left, res, left -= root->val); 22 if(root->right) 23 dfs(root->right, res, left -= root->val); 24 } 25 } 26 private: 27 vector<vector<int>> ret; 28 };
java版本代码如下，方法相同，就是java的引用处理起来稍微麻烦点，递归尾部应该pop一下。
1 public class Solution { 2 public List<List<Integer>> pathSum(TreeNode root, int sum) { 3 List<List<Integer>> ret = new ArrayList<List<Integer>>(); 4 Stack<Integer> tmp = new Stack<Integer>(); 5 dfs(ret, tmp, root, sum); 6 return ret; 7 } 8 9 public void dfs(List<List<Integer>> ret, Stack<Integer> tmp, TreeNode root, int remain){ 10 if(root == null) return; 11 tmp.push(root.val); 12 if(root.left == null && root.right == null) 13 if(remain == root.val) 14 ret.add((Stack<Integer>)tmp.clone()); 15 if(root.left != null) dfs(ret, tmp, root.left, remain - root.val); 16 if(root.right != null) dfs(ret, tmp, root.right, remain - root.val); 17 tmp.pop(); 19 } 20 }