Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum.
For example:sum
= 22
,
5 / 4 8 / / 11 13 4 / \ / 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
解题思路:
深度优先递归搜索。注意节点的值可能是负数。
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int>> pathSum(TreeNode* root, int sum) { vector<vector<int>> result; vector<int> resultItem; helper(result, resultItem, root, sum); return result; } void helper(vector<vector<int>>& result, vector<int> resultItem, TreeNode* root, int leftSum){ if(root==NULL){ return; } if(root->left==NULL && root->right==NULL){ //root为叶子节点 if(leftSum==root->val){ resultItem.push_back(root->val); result.push_back(resultItem); } }else{ resultItem.push_back(root->val); leftSum = leftSum - root->val; if(root->left!=NULL){ helper(result, resultItem, root->left, leftSum); } if(root->right!=NULL){ helper(result, resultItem, root->right, leftSum); } } } };
原文地址:http://blog.csdn.net/kangrydotnet/article/details/46548257