标签：breadth-first search leetcode depth-first search
题目：
Follow up for problem "Populating Next Right Pointers in Each Node".
What if the given tree could be any binary tree? Would your previous solution still work?
Note:
For example,
Given the following binary tree,
1
/ 2 3
/ \ 4 5 7
After calling your function, the tree should look like:
1 -> NULL
/ 2 -> 3 -> NULL
/ \ 4-> 5 -> 7 -> NULL
思路：这次我们要链接的不是完全二叉树，题目依然要求我们使用常数空间复杂度。我们需要多维护两个变量，总共四个变量。
lasthead = curhead; //移动到下一层 curhead = NULL; //重置curhead;
/**
* Definition for binary tree with next pointer.
* struct TreeLinkNode {
* int val;
* TreeLinkNode *left, *right, *next;
* TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void connect(TreeLinkNode *root) {
if(root == NULL) return;
TreeLinkNode* lasthead = root; //维护上一层的链表开始的节点
TreeLinkNode* pre = NULL; //维护前一个结点
TreeLinkNode* curhead = NULL; //维护当前层的链表开始的节点
while(lasthead)
{
TreeLinkNode* lastCur = lasthead; //lastcur从上一层链表的开始节点开始
while(lastCur)
{
if(lastCur->left)
{
if(curhead == NULL)
{
curhead = lastCur->left; //如果还没确定该行的头结点，在碰到第一个非空节点时确定
pre = curhead;
}
else
{
pre->next = lastCur->left;
pre = pre->next;
}
}
if(lastCur->right)
{
if(curhead == NULL)
{
curhead = lastCur->right;
pre = curhead;
}
else
{
pre->next = lastCur->right;
pre = pre->next;
}
}
lastCur = lastCur->next; //移动到下一个节点
}
lasthead = curhead; //移动到下一层
curhead = NULL; //重置curhead;
}
}
};[C++]LeetCode: 124 Populating Next Right Pointers in Each Node II(链接二叉树)
标签：breadth-first search leetcode depth-first search
原文地址：http://blog.csdn.net/cinderella_niu/article/details/43151057
