[LintCode] 246 Binary Tree Path Sum II 解题报告

Description
Your are given a binary tree in which each node contains a value. Design an algorithm to get all paths which sum to a given value. The path does not need to start or end at the root or a leaf, but it must go in a straight line down.




Example
Given a binary tree:

    1
   / \
  2   3
 /   /
4   2
for target = 6, return

[
  [2, 4],
  [1, 3, 2]
]



思路
题目要求不一定从root出发，但是一定要从上往下。
我们还是从root出发往下traverse。并且维护一边往下走一边把遍历到的点的值放进一个list里。
在当前点，我们做一件事：
把当前的list从后往前加，看能不能加到target。
如果加到target，说明有一个解。我们把这个解放进result里。
全部check完毕以后，我们可以继续往下走了。
最后全部traverse结束，我们返回答案的List。




Code

/**
 * 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) {
        // Write your code here
        
        List<List<Integer>> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        
        List<Integer> list = new ArrayList<>();
        helper(root, result, list, target);
        return result;
    }
    
    public void helper(TreeNode root, List<List<Integer>> result, List<Integer> list, int target) {
        
        if (root == null) {
            return;
        }
        
        list.add(root.val);
        int sum = 0;
        for (int i = list.size() - 1; i >= 0; i--) {
            sum += list.get(i);
            if (sum == target) {
                List<Integer> res = new ArrayList<>();
                for (int j = i; j < list.size(); j++) {
                    res.add(list.get(j));
                }
                result.add(res);
            }
        }
        
        helper(root.left, result, list, target);
        helper(root.right, result, list, target);
        list.remove(list.size() - 1);
    }
}