[leedcode 113] Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / 
            4   8
           /   / 
          11  13  4
         /      / 
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
        List<List<Integer>> res;
        List<Integer> seq;
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        //本题求的是中间结果,需要构造一个函数，函数参数sum表示sum-当前已经遍历过的和，递归终止条件：当遍历到叶子节点，
        //和为0时，添加结果。注意遍历完左节点和右节点之后，不要忘记弹出最近放入的节点
        res=new ArrayList<List<Integer>>();
        seq=new ArrayList<Integer>();
        if(root==null) return res;
        getPath(root,sum);

        return res;
    }
    public void getPath(TreeNode root,int sum){
        if(root==null) return;
        seq.add(root.val);
        if(root.left==null&&root.right==null&&(sum-root.val==0)){
            res.add(new ArrayList<Integer>(seq));
        }
        getPath(root.left,sum-root.val);
        getPath(root.right,sum-root.val);
        seq.remove(seq.size()-1);////***
    }
}