路径总和 II 给你二叉树的根节点 root 和一个整数目标和 targetSum ,找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。
叶子节点 是指没有子节点的节点。
示例 1:
输入:root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22 输出:[[5,4,11,2],[5,8,4,5]] 示例 2:
输入:root = [1,2,3], targetSum = 5 输出:[] 示例 3:
输入:root = [1,2], targetSum = 0 输出:[]
提示:
树中节点总数在范围 [0, 5000] 内 -1000 <= Node.val <= 1000 -1000 <= targetSum <= 1000
/**
* 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 {
vector
void BackTracking(TreeNode* root, int targetSum,int Sum,vector
{
if(root->left==nullptr&&root->right==nullptr&&Sum==targetSum)
{
result.push_back(path);
return;
}
if(root->left)
{
int num=root->left->val;
path.push_back(num);
Sum+=num;
BackTracking(root->left,targetSum,Sum,path);
path.pop_back();
Sum-=num;
}
if(root->right)
{
int num=root->right->val;
path.push_back(num);
Sum+=num;
BackTracking(root->right,targetSum,Sum,path);
path.pop_back();
Sum-=num;
}
}
public:
vector
if(root==nullptr)
return result;
vector
path.push_back(root->val);
BackTracking(root,targetSum,root->val,path);
return result;
}
};
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