标签：style blog java color os 2014
题目：给定一颗完全二叉树，连接每个节点的next为该节点的右边同一层的节点
例如：
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
算法：
将根节点压入队列中 若队列不为空，则执行循环 取出并弹出队列首节点 if 队列不为空 then if 当前节点和队列队首节点位于二叉树的同一层 then current_node.next = next_node end end 将当前节点的左孩子和右孩子压入队尾
/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
* int val;
* TreeLinkNode left, right, next;
* TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
/**
* Algorithm:
*
* 1. push root into queue
* while queue is not empty
* 2. get and pop the top tree node from queue
* 3. if queue has any tree node more
* 4. if current_node and next_node at the same depth
* 5. then current_node.next = next_node
* 6. push current.node.left and current_node.right into queue
*
*/
public void connect(TreeLinkNode root) {
if (null == root) {
// Empty Tree
return ;
}
int counter = 0; // means the ith node of the tree
Queue<TreeLinkNode> queue = new LinkedList<TreeLinkNode>();
queue.add(root);
while (!queue.isEmpty()) {
TreeLinkNode node = queue.poll();
++counter;
if (!queue.isEmpty()) {
int curDepth = (int)(Math.log(1.0*counter)/Math.log(2.0));
int nextDepth = (int)(Math.log(1.0*counter+1.0)/Math.log(2.0));
if (curDepth == nextDepth) {
TreeLinkNode nextNode = queue.peek();
node.next = nextNode;
}
}
if (null != node.left) {
queue.add(node.left);
}
if (null != node.right) {
queue.add(node.right);
}
}
}
}[LeetCode]Populating Next Right Pointers in Each Node,布布扣,bubuko.com
[LeetCode]Populating Next Right Pointers in Each Node
标签：style blog java color os 2014
原文地址：http://blog.csdn.net/yeweiouyang/article/details/37818921
