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]
]
这题和Path Sum I一样使用前序遍历的方法，不过遍历的同时需要保存当前的值。
设当前访问的节点为p，已访问的节点保存于cur, 结果集result, 目标值sum
- 若p是空节点，直接返回
- p值压入cur，
sum -= p->val - 若p是叶子节点，并且
sum == 0, 把cur压入result结果集中 - 否则p不是叶子节点，递归探测左孩子、右孩子.
void pathSum(vector<vector<int>> &result, vector<int> cur, TreeNode *root, int sum) {
if (root == NULL)
return;
cur.push_back(root->val);
sum -= root->val;
if (isLeft(root) && sum == 0) {
result.push_back(cur);
return;
}
pathSum(result, cur, root->left, sum);
pathSum(result, cur, root->right, sum);
}