Populating Next Right Pointers in Each Node
描述
Given a binary tree
struct TreeLinkNode {
int val;
TreeLinkNode *left, *right, *next;
TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
};
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,
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
分析
无
代码
// Populating Next Right Pointers in Each Node
// 时间复杂度O(n),空间复杂度O(logn)
public class Solution {
public void connect(TreeLinkNode root) {
connect(root, null);
}
private static void connect(TreeLinkNode root, TreeLinkNode sibling) {
if (root == null) return;
else root.next = sibling;
connect(root.left, root.right);
if (sibling != null) connect(root.right, sibling.left);
else connect(root.right, null);
}
}