标签：end from sam ali param class tco turn 方法
Given a binary tree, find all paths that sum of the nodes in the path equals to a given number target.
A valid path is from root node to any of the leaf nodes.
Example 1:
Input:
{1,2,4,2,3}
5
Output: [[1, 2, 2],[1, 4]]
Explanation:
The tree is look like this:
1
/ 2 4
/ 2 3
For sum = 5 , it is obviously 1 + 2 + 2 = 1 + 4 = 5
Example 2:
Input: {1,2,4,2,3} 3 Output: [] Explanation: The tree is look like this: 1 / 2 4 / 2 3 Notice we need to find all paths from root node to leaf nodes. 1 + 2 + 2 = 5, 1 + 2 + 3 = 6, 1 + 4 = 5 There is no one satisfying it.
思路：
递归法，用helper方法，可以传更多参数
注意：
代码：
/* * @param root: the root of binary tree * @param target: An integer * @return: all valid paths */ public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) { List<List<Integer>> result = new ArrayList<>(); ArrayList<Integer> path = new ArrayList<>(); if (root == null) { return result; } path.add(root.val); helper(root, path, root.val, target, result); return result; } public void helper (TreeNode root, ArrayList<Integer> path, int sum, int target, List<List<Integer>> result) { //meet leaf if (root.right == null && root.left == null) { if (sum == target) { result.add(new ArrayList(path)); } return; } // right node if (root.right != null) { path.add(root.right.val); helper(root.right, path, sum + root.right.val, target, result); path.remove(path.size() - 1); } //left node if (root.left != null) { path.add(root.left.val); helper(root.left, path, sum + root.left.val, target, result); path.remove(path.size() - 1); } }
Lintcode376-Binary Tree Path Sum-Easy
标签：end from sam ali param class tco turn 方法
原文地址：https://www.cnblogs.com/Jessiezyr/p/10656726.html
