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)
public class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> result = new ArrayList<>();
ArrayList<Integer> cur = new ArrayList<>(); // 中间结果
pathSum(root, sum, cur, result);
return result;
}
private static void pathSum(TreeNode root, int gap, ArrayList<Integer> cur,
List<List<Integer>> result) {
if (root == null) return;
cur.add(root.val);
if (root.left == null && root.right == null) { // leaf
if (gap == root.val)
result.add(new ArrayList<>(cur));
}
pathSum(root.left, gap - root.val, cur, result);
pathSum(root.right, gap - root.val, cur, result);
cur.remove(cur.size() - 1);
}
}