实现思路
对树进行前序遍历,在遍历过程中,保存路径和累加和,当遍历到根结点时也就是左右结点为nullptr的点时,判断加上该结点的sum是否和targetNum相等,如果相等将路径保存到结果的vector中
实现代码
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
void solve(TreeNode* root,vector<int> path,int sum,int target,vector<vector<int>> &re){
if(root==nullptr) return;
if(root->left==nullptr&&root->right==nullptr){
sum+=root->val;
path.push_back(root->val);
if(sum==target)
{
re.push_back(path);
}
return;
}
sum+=root->val;
path.push_back(root->val);
solve(root->left,path,sum,target,re);
solve(root->right,path,sum,target,re);
}
vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
vector<vector<int>> re;
vector<int> path;
solve(root,path,0,targetSum,re);
return re;
}
};
需要掌握的代码:
#include < algorithm >
vector< int >path=[3,4,1,2];
reverse(path.begin(),path.end());
提交结果及分析
时间复杂度
O(n)