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The problem:
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:
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
First solution : <O(n) space complexity>
public class Solution { public void connect(TreeLinkNode root) { if (root == null) return; Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode> (); TreeLinkNode cur_node; int cur_num = 1; int next_num = 0; queue.offer(root); while (queue.size() != 0) { cur_node = queue.poll(); cur_num--; if (cur_node.left != null) { queue.offer(cur_node.left); next_num++; } if (cur_node.right != null) { queue.offer(cur_node.right); next_num++; } if (cur_num != 0) { cur_node.next = queue.peek(); } else{ cur_node.next = null; cur_num = next_num; next_num = 0; } } } }
[LeetCode#]Populating Next Right Pointers in Each Node
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原文地址：http://www.cnblogs.com/airwindow/p/4235825.html
