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[LeetCode]116.Populating Next Right Pointers in Each Node
【题目】
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
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【分析】
在层次遍历过程中，修改next指针指向。
类似于： [LeetCode]Binary Tree Level Order Traversal
【代码】
/*********************************
* 日期：2014-12-24
* 作者：SJF0115
* 题目: 116.Populating Next Right Pointers in Each Node
* 来源：https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/
* 结果：AC
* 来源：LeetCode
* 总结：
**********************************/
#include <iostream>
#include <queue>
using namespace std;
struct TreeLinkNode {
int val;
TreeLinkNode *left;
TreeLinkNode *right;
TreeLinkNode *next;
TreeLinkNode(int x):val(x),left(NULL),right(NULL),next(NULL){}
};
class Solution {
public:
void connect(TreeLinkNode *root) {
if(root == NULL){
return;
}//if
queue<TreeLinkNode*> cur;
queue<TreeLinkNode*> next;
cur.push(root);
TreeLinkNode *p,*pre;
while(!cur.empty()){
pre = NULL;
// 当前层遍历
while(!cur.empty()){
// 出队列
p = cur.front();
cur.pop();
// 横向连接
if(pre != NULL){
pre->next = p;
}//if
pre = p;
// next保存下一层节点
// 左子树不空加入队列
if(p->left){
next.push(p->left);
}//if
// 右子树不空加入队列
if(p->right){
next.push(p->right);
}//if
}//while
p->next = NULL;
swap(next,cur);
}//while
}
};
//按先序序列创建二叉树
int CreateBTree(TreeLinkNode*& T){
int data;
//按先序次序输入二叉树中结点的值,-1表示空树
cin>>data;
if(data =http://www.mamicode.com/= -1){>
[LeetCode]116.Populating Next Right Pointers in Each Node
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