Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references. A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.
Example1:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: [[5,4,11,2],[5,8,4,5]]
Explanation: There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22
Example2:
Input: root = [1,2,3], targetSum = 5
Output: []
Example3:
Input: root = [1,2], targetSum = 0
Output: []
Constraints:
- The number of nodes in the tree is in the range [0, 5000].
- -1000 <= Node.val <= 1000
- -1000 <= targetSum <= 1000
这道题其实和Path Sum很相似。不同的是需要把path给记下来。其实知道题就是dfs(深度优先搜索), 每次都要到叶子结点满足条件才算是真的满足条件。
注意:
- DFS中每次回到前面的结点都要把之前加入的那个值去掉,不然的话path就不对了
- 把path加到最后返回的list的时候,需要做一次深度copy,不然的话path改变,结果的list也跟着改变。
- 深度copy的时候怎么写
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
List<List<Integer>> rst = new ArrayList<>();
if (root == null) {
return rst;
}
List<Integer> path = new ArrayList<>();
process(root, targetSum, rst, path);
return rst;
}
private void process(TreeNode root, int targetSum,
List<List<Integer>> rst, List<Integer> path) {
if (root == null) {
return;
}
path.add(root.val);
if (root.left == null && root.right == null && root.val == targetSum) {
rst.add(new ArrayList(path));
}
process(root.left, targetSum - root.val, rst, path);
process(root.right, targetSum - root.val, rst, path);
path.remove(path.size() - 1);
}
}