标签：blog win 循环 ext calling turn linked div binary
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
bfs + 记住每层的前一个节点像是linkedlist
/**
* 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;
Queue<TreeLinkNode> queue = new LinkedList<>();
queue.offer(root);
TreeLinkNode pre = null;
while (!queue.isEmpty()) {
int size = queue.size();
for (int i=0; i<size; i++) {
TreeLinkNode cur = queue.poll();
if (i == 0) pre = cur;
else {
pre.next = cur;
pre = cur;
}
if (cur.left != null) queue.offer(cur.left);
if (cur.right != null) queue.offer(cur.right);
}
}
}
}
双循环(通过next) : 外层遍历(记住下层节点的起始位置left)- 向下遍历--
内循环(通过cur.next)-- 向右遍历本层节点构造next;
public class Solution {
public void connect(TreeLinkNode root) {
if(root==null) return;
root.next = null;
TreeLinkNode cur = root;
TreeLinkNode nextLeftmost = null;
//递进while(cur.left!=null){
//递进nextLeftmost = cur.left; // save the start of next level, 当作下次内循环的开始点并判断是否有下次循环
//层次 while(cur!=null){
cur.left.next=cur.right;
cur.right.next = cur.next==null? null : cur.next.left;
cur=cur.next;
}
cur=nextLeftmost; // point to next level
}
}
}
116. Populating Next Right Pointers in Each Node
标签：blog win 循环 ext calling turn linked div binary
原文地址：http://www.cnblogs.com/apanda009/p/7295592.html
