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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
/** * 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) { //利用使用一个队列 Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>(); int level = 1;//记录需要出队的元素数, if(root == null ){ return; } queue.offer(root); int count = 0; TreeLinkNode p = null; while(queue.size() > 0){ while(count < level - 1){ p = queue.poll(); p.next = queue.peek(); if(p.left != null && p.right != null){ queue.offer(p.left); queue.offer(p.right); } count++; } p = queue.poll(); if(p.left != null && p.right != null){ queue.offer(p.left); queue.offer(p.right); } level = level * 2; count = 0; } } }
题意就是将同层节点连成链,所以想到的是BFS搜索,使用一个队列存储,由于是完全二叉树,因此可以使用level来方便的控制出队的个数,从而也就处理好了层次的问题。
Runtime: 253 ms
Populating Next Right Pointers in Each Node
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