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给一棵二叉树和一个目标值,设计一个算法找到二叉树上的和为该目标值的所有路径。路径可以从任何节点出发和结束,但是需要是一条一直往下走的路线。也就是说,路径上的节点的层级是逐个递增的。
样例
对于二叉树:
1
/ \
2 3
/ /
4 2
给定目标值6
。那么满足条件的路径有两条:
[
[2, 4],
[1, 3, 2]
]
解题思路:
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/*
* @param root: the root of binary tree
* @param target: An integer
* @return: all valid paths
*/
public List<List<Integer>> binaryTreePathSum2(TreeNode root, int target) {
List<List<Integer>> list = new ArrayList<>();
if (root == null) return list;
helper(root, target, list, new ArrayList<Integer>());
return list;
}
private void helper(TreeNode root, int target, List<List<Integer>> list, List<Integer> path) {
if (root == null) return;
int sum = 0;
path.add(Integer.valueOf(root.val));
for (int i = path.size() - 1; i >= 0; i--) {
sum += path.get(i);
if (sum == target)
list.add(new ArrayList<Integer>(path.subList(i, path.size())));
}
if (root.left != null)
helper(root.left, target, list, path);
if (root.right != null)
helper(root.right, target, list, path);
path.remove(path.size() - 1);
}
}