Path for a binary tree and value - prove safety offer

An input binary tree root node and an integer, the binary print values ​​of nodes in the input path and all integers. Forming a path to the path definition begins from the root node of the tree down to the leaf node has been traversed nodes. (Note: the return value in the list, a large array Array front)

A preorder traversal of the tree to record the path

public class Solution {
    ArrayList<ArrayList<Integer>> paths=new ArrayList<ArrayList<Integer>>();
    ArrayList<Integer> path=new ArrayList<Integer>();//存储当前的路径
    public ArrayList<ArrayList<Integer>> FindPath(TreeNode root,int target) {
        if(target < 0){
            return  paths;
        }
        if(root == null){
            return paths;
        }
        path.add(root.val);
        if(root.left == null &&root.right==null&& target == root.val){
            paths.add(new ArrayList<Integer>(path));
        }
        if(root.val < target && root.left!=null){
            FindPath(root.left,target-root.val);
        }
        if(root.val < target&& root.left!=null){
            FindPath(root.right,target-root.val);
        }
        //遍历完一条路径之后需要回溯
        path.remove(path.size()-1);
        return paths;
    }
}

Reference blog: https://blog.csdn.net/u014525494/article/details/80978647

After only fully understood in view of a part of the recording, etc.