113. 路径总和 II
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2022-05-20 11:21:25
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给定一个二叉树和一个目标和,找到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。
说明: 叶子节点是指没有子节点的节点。
示例:
给定如下二叉树,以及目标和 sum = 22
,
5 / \ 4 8 / / \ 11 13 4 / \ / \ 7 2 5 1
返回:
[ [5,4,11,2], [5,8,4,5] ]
/**
* 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;
if(!root){
return result;
}
vector<int> answer;
backtrace(root, sum, answer, result);
return result;
}
void backtrace(TreeNode* root, int sum, vector<int> answer, vector<vector<int>>& result){
//assert(root);
if(root && root->val == sum && !root->left && !root->right){
answer.push_back(root->val);
result.push_back(answer);
}
answer.push_back(root->val);
if(root->left){
backtrace(root->left, sum-root->val, answer, result);
}
if(root->right){
backtrace(root->right, sum-root->val, answer, result);
}
answer.pop_back();
}
};