Given a binary tree
struct TreeLinkNode {
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *next;
}
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
Initially, all next pointers are set to NULL.
Note:
- You may only use constant extra space.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,
1
/
2 3
/ /
4 5 6 7
After calling your function, the tree should look like:
1 -> NULL
/
2 -> 3 -> NULL
/ /
4->5->6->7 -> NULL
题目含义：给定一个完全二叉树，从左到右连接每一层节点.
完全二叉树：所有叶子都在同一深度，每一个根节点都有两个叶子节点
方法一：
1 public void connect(TreeLinkNode root) { 2 // 由于是完全二叉树，所以若节点的左子结点存在的话，其右子节点必定存在，所以左子结点的next指针可以直接指向其右子节点，对于其右子节点的处理方法是，判断其父节点的next是否为空，若不为空，则指向其next指针指向的节点的左子结点，若为空则指向NULL 3 if (root ==null ) return; 4 if (root.left != null) root.left.next = root.right; 5 if (root.right!=null) root.right.next = root.next!=null?root.next.left :null; 6 connect(root.left); 7 connect(root.right); 8 }
方法二：会超时
1 public void connect(TreeLinkNode root) { 2 List<Integer> result = new ArrayList<>(); 3 if (root == null) return; 4 Queue<TreeLinkNode> q = new LinkedList<>(); 5 q.add(root); 6 while (!q.isEmpty()) { 7 int size = q.size(); 8 TreeLinkNode prev = null; 9 TreeLinkNode cur = null; 10 for (int i = 0; i < size - 1; i++) { 11 cur = q.poll(); 12 if (prev == null) { 13 prev = cur; 14 } else { 15 prev.next = cur; 16 prev = cur; 17 } 18 19 if (cur.left != null) q.offer(cur.left); 20 if (cur.right != null) q.offer(cur.right); 21 } 22 } 23 }