题目：
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]
]
题解：
与上一题不同的是，要存储所有的root - leaf路径，所以在DFS的基础上我们还要增加backtracking。回溯的点有两个，一个是当前root满足条件，返回时，此时回溯保证当前root的sibling结果正确； 另外一个是遍历完当前root的左子树和右子树时， 这时回溯也是保证sibling结果正确。比如 root = [0, 1, 1], sum = 1，则计算完左边的1时可以正确计算到右边的1。或者 root = [0, 1, 1, 0, 0]， sum = 1， 遍历完左边的两个0后可以正确计算到右边1的结果。 (有空要组织一下语言)
Time Complexity - O(n)， Space Complexity - O(n)。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> res = new ArrayList<>();
ArrayList<Integer> list = new ArrayList<>();
dfs(res, list, root, sum);
return res;
} private void dfs(List<List<Integer>> res, ArrayList<Integer> list, TreeNode root, int sum) {
if(root == null)
return;
if(root.left == null && root.right == null && root.val == sum) {
list.add(root.val);
res.add(new ArrayList<Integer>(list));
list.remove(list.size() - 1);
return;
} list.add(root.val);
sum -= root.val;
dfs(res, list, root.left, sum);
dfs(res, list, root.right, sum);
list.remove(list.size() - 1);
}
}
二刷:
注意回溯的点。
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> res = new ArrayList<>();
List<Integer> list = new ArrayList<>();
pathSum(res, list, root, sum);
return res;
} private void pathSum(List<List<Integer>> res, List<Integer> list, TreeNode root, int sum) {
if (root == null) return;
sum -= root.val;
list.add(root.val);
if (root.left == null && root.right == null && sum == 0) {
res.add(new ArrayList<>(list));
list.remove(list.size() - 1);
return;
}
pathSum(res, list, root.left, sum);
pathSum(res, list, root.right, sum);
list.remove(list.size() - 1);
}
}
测试：