113. 路径总和 II
力扣链接(中等):https://leetcode.cn/problems/path-sum-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:

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| 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]]
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示例 2:

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| 输入:root = [1,2,3], targetSum = 5
输出:[]
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示例 3:
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| 输入:root = [1,2], targetSum = 0
输出:[]
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提示:
- 树中节点总数在范围
[0, 5000]
内
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
个人题解
C++ |
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| /**
* 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:
vector<vector<int>> result;
vector<int> path;
void traversal(TreeNode* node, int count) {
// 处理叶子节点
if(!node->left && !node->right && count == 0) {
result.push_back(path);
return;
}
if(!node->left && !node->right) return;
// 递归
if(node->left) {
path.push_back(node->left->val);
count -= node->left->val;
traversal(node->left, count);
count += node->left->val;
path.pop_back();
}
if(node->right) {
path.push_back(node->right->val);
count -= node->right->val;
traversal(node->right, count);
count += node->right->val;
path.pop_back();
}
}
vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
if(!root) return vector<vector<int>>();
path.push_back(root->val);
traversal(root, targetSum - root->val);
return result;
}
};
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