Path Sum II
描述
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
跟上一题相比，本题是求路径本身。且要求出所有结果，左子树求到了满意结果，不能return，要接着求右子树。
代码
// Path Sum II
// 时间复杂度O(n)，空间复杂度O(logn)
class Solution {
public:
vector<vector<int> > pathSum(TreeNode *root, int sum) {
vector<vector<int> > result;
vector<int> cur; // 中间结果
pathSum(root, sum, cur, result);
return result;
}
private:
void pathSum(TreeNode *root, int gap, vector<int> &cur,
vector<vector<int> > &result) {
if (root == nullptr) return;
cur.push_back(root->val);
if (root->left == nullptr && root->right == nullptr) { // leaf
if (gap == root->val)
result.push_back(cur);
}
pathSum(root->left, gap - root->val, cur, result);
pathSum(root->right, gap - root->val, cur, result);
cur.pop_back();
}
};