leetcode_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.

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

return

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

public List<List<Integer>> pathSum(TreeNode root, int sum){
		List<List<Integer>>list=new ArrayList<List<Integer>>();
		int num=0;
		if (root == null)
			return list;
		List<Integer>subList=new ArrayList<Integer>();
		LinkedList<TreeNode>linkedList=new LinkedList<TreeNode>();
		Stack<TreeNode> st1 = new Stack<TreeNode>();
		linkedList.addLast(root);
		num+=root.val;
		TreeNode top = null;
		while (!linkedList.isEmpty()) {
			top=linkedList.getLast();
			while(top.left!=null)//达到页结点
			{
				linkedList.addLast(top.left);
				num+=top.left.val;
				top=top.left;
			}
			while(!linkedList.isEmpty())
			{
				top=linkedList.getLast();
				if(top.right==null)//无右子树时，访问结点
				{
					if(top.left==null&&num==sum)
					{
						for(int i=0;i<linkedList.size();i++)
							subList.add(linkedList.get(i).val);
						list.add(subList);
						subList=new ArrayList<Integer>();
					}
					linkedList.removeLast();
					num-=top.val;
				}else 
				{
					if(st1.empty()||(!st1.empty()&&st1.peek()!=top))//有右子树且第一次出现，将右子树入栈，并将结点在st1中标记一下
					{
						st1.push(top);
						linkedList.addLast(top.right);
						num+=top.right.val;
						break;//右子树入栈了，然后从右子树开始
					}
					else//有右子树，但是第二次出现了，访问该结点
					{
						st1.pop();
						linkedList.removeLast();
						num-=top.val;
					}
				}
				
			}
		}
		return list;
    }