Populating Next Right Pointers in Each Node
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
思想:递归,左子树和右子树,以及之间的关联。要与基本的先序、中序和后序联系起来
java代码:
- public void connect(TreeLinkNode root) {
- if(root==null) return;
- if(root.left == null && root.right == null) return;
- connect(root.left);
- connect(root.right);
- TreeLinkNode left = root.left;
- TreeLinkNode right = root.right;
- while(left!=null) {
- left.next = right;
- left = left.right;
- right = right.left;
- }
- }