标签:
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]
]
分析:
这道题目算是Path Sum的延伸,只不过要求出所有从根到树叶的路径上节点和为给定值的路径,并放到一个vector中返回。采取类似于Path Sum递归思路,每次递归都会建立一个新的vector,复制之前的vector内容,并将当前节点的值放入,如果当前节点是叶节点,并且路径上节点和满足要求,就将新的vector放入最终返回的vector中。
代码:
class Solution {
public:
vector<vector<int>> res;
vector<vector<int>> pathSum(TreeNode* root, int sum) {
if(!root)
return res;
vector<int> tmp;
pathSum1(root,sum,tmp);
return res;
}
void pathSum1(TreeNode* root, int sum,vector<int> tmp)
{
if(!root)
return ;
vector<int> tmp1;
tmp1 = tmp;
tmp1.push_back(root->val);
if(!root->left&&!root->right)
{
if(root->val==sum)
res.push_back(tmp1);
return;
}
pathSum1(root->left,sum-root->val,tmp1);pathSum1(root->right,sum-root->val,tmp1);
}
};
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原文地址:http://blog.csdn.net/pwiling/article/details/50957763