[leetcode]113. Path Sum II

backtrack的套路{
先加这一层t
判断是否满足，满足就输出
递归继续其他的选择t+1
减掉这一层t
}
可参见40题

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> res=new ArrayList<>();
        backtrack(root,sum,0,res,new ArrayList<>());
        return res;
    }
    
    public void backtrack(TreeNode root,int sum,int curr,List<List<Integer>> res, List<Integer> list){
        if(root==null) return;
        
        curr+=root.val;
        list.add(root.val);
        
        if(curr==sum&&root.left==null&&root.right==null){ //这里一定要加上叶子限制的条件
            res.add(new ArrayList<Integer>(list));
        }
        backtrack(root.left,sum,curr,res,list);
        backtrack(root.right,sum,curr,res,list);
        
        curr-=root.val;
        list.remove(list.size()-1);        
        
    }
}