Java for LeetCode 117 Populating Next Right Pointers in Each Node II

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

• You may only use constant extra space.

For example,
Given the following binary tree,

         1
       /  \
      2    3
     / \    \
    4   5    7

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL

 

解题思路一：

本题的难点在于需要逐层遍历才行，因此可以用Java for LeetCode 102 Binary Tree Level Order Traversal的思路，JAVA实现如下：

    public void connect(TreeLinkNode root) {
    	if (root == null || (root.left == null && root.right == null))
			return;
    	List<List<TreeLinkNode>> list=new ArrayList<List<TreeLinkNode>>();
        Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
        queue.add(root);
        while (queue.size() != 0) {
            List<TreeLinkNode> alist = new ArrayList<TreeLinkNode>();
            for (TreeLinkNode child : queue)
                alist.add(child);
            list.add(new ArrayList<TreeLinkNode>(alist));
            Queue<TreeLinkNode> queue2=queue;
            queue=new LinkedList<TreeLinkNode>();
            for(TreeLinkNode child:queue2){
                if (child.left != null)
                    queue.add(child.left);
                if (child.right != null)
                    queue.add(child.right);
            }
        }
        for(List<TreeLinkNode> alist:list)
        	for(TreeLinkNode aNode:alist)
        		connectARoot(aNode);
    }
    public static void connectARoot(TreeLinkNode root){
    	if (root == null || (root.left == null && root.right == null))
			return;
		if (root.next == null) {
			if (root.left != null)
				root.left.next = root.right;
		}
		else if (root.right == null) {
			TreeLinkNode temp=root.next;
			while(temp!=null){
				if(temp.left==null&&temp.right==null)
					temp=temp.next;
				else break;
			}
			if(temp!=null)
				root.left.next = (temp.left == null?temp.right:temp.left);
		}else {
			if (root.left != null)
				root.left.next = root.right;
			TreeLinkNode temp=root.next;
			while(temp!=null){
				if(temp.left==null&&temp.right==null)
					temp=temp.next;
				else break;
			}
			if(temp!=null)
				root.right.next = (temp.left == null?temp.right:temp.left);
		}
    }

解题思路二：

不使用队列，请移步Populating Next Right Pointers in Each Node I II@LeetCode