Leetcode: Populating Next Right Pointers in Each Node
2015-01-28 07:19
288 查看
Problem:
Given a binary tree
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
Initially, all next pointers are set to
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,
After calling your function, the tree should look like:
Solution:
Idea: Pre-order traverse.
Code:
/**
* 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 || (root.left == null && root.right == null)) return;
root.left.next = root.right;
if (root.next != null) root.right.next = root.next.left;
connect(root.right);
connect(root.left);
}
}
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
Solution:
Idea: Pre-order traverse.
Code:
/**
* 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 || (root.left == null && root.right == null)) return;
root.left.next = root.right;
if (root.next != null) root.right.next = root.next.left;
connect(root.right);
connect(root.left);
}
}
相关文章推荐
- [LeetCode] Populating Next Right Pointers in Each Node
- [Leetcode] Populating Next Right Pointers in Each Node
- [Leetcode] Populating Next Right Pointers in Each Node ii
- LeetCode: Populating Next Right Pointers in Each Node
- 【leetcode】Populating Next Right Pointers in Each Node II
- LeetCode Populating Next Right Pointers in Each Node
- LeetCode - Populating Next Right Pointers in Each Node
- Leetcode: Populating Next Right Pointers in Each Node
- leetcode 79: Populating Next Right Pointers in Each Node II
- LeetCode: Populating Next Right Pointers in Each Node
- [LeetCode] Populating Next Right Pointers in Each Node
- LeetCode(Oct28'12):Populating Next Right Pointers in Each Node II
- [LeetCode]Populating Next Right Pointers in Each Node
- leetcode 78: Populating Next Right Pointers in Each Node
- [LeetCode] Populating Next Right Pointers in Each Node II
- [LeetCode] Populating Next Right Pointers in Each Node II 解题报告
- 【leetcode】 Populating Next Right Pointers in Each Node
- leetcode Populating Next Right Pointers in Each Node II
- [Leetcode]Populating Next Right Pointers in Each Node II
- leetcode 79: Populating Next Right Pointers in Each Node II