LC116. 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
利用递归，把每一棵树的左右连起来。由于是完美二叉树，所以可以直接连接左边的右和右边的左。
/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
* int val;
* TreeLinkNode left, right, next;
* TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
public void connect(TreeLinkNode root) {
if(root == null) return;
connect(root.left, root.right);
}
public void connect(TreeLinkNode node1, TreeLinkNode node2){
if(node1 == null){
return;
}
node1.next = node2;
if(node1.left != null){
connect(node1.left, node1.right);
connect(node2.left, node2.right);
connect(node1.right, node2.left);
}
}
}
O(n), O(height)
LC117. 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
常数空间，就不能用递归啦。
每层进行处理，将一层的节点都链接好才进入下一层。
/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
* int val;
* TreeLinkNode left, right, next;
* TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
public void connect(TreeLinkNode root) {
// level by level link node
TreeLinkNode temp = new TreeLinkNode(0);
TreeLinkNode cur = null;
while(root != null){
cur = temp;
// link current level
while(root != null){
if(root.left != null){
cur.next = root.left;
cur = cur.next;
}
if(root.right != null){
cur.next = root.right;
cur = cur.next;
}
root = root.next;
}
// next level
root = temp.next;
temp.next = null;
} // while
return;
}
}
O(n), O(1).