116. Populating Next Right Pointers in Each Node - Medium
阿新 • • 發佈:2019-01-01
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.
- Recursive approach is fine, implicit stack space does not count as extra space for this problem.
- You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
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
用兩個指標,start和cur,start標記每一層的起點(最左),cur遍歷該層,逐層遍歷,start從root開始
time: O(n), space: O(1)
/** * 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; } TreeLinkNode start = root, cur = null; while(start.left != null) { cur = start; while(cur != null) { cur.left.next = cur.right; if(cur.next != null) { cur.right.next = cur.next.left; } cur = cur.next; } start = start.left; } } }