[LeetCode]117.Populating Next Right Pointers in Each Node II

标签：层次遍历   leetcode   二叉树   

【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:

• You may only use constant extra space.

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

【分析】

无

【代码】

/*********************************
*   日期：2014-12-24
*   作者：SJF0115
*   题目: 117.Populating Next Right Pointers in Each Node II
*   来源：https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/
*   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 == -1){
        T = NULL;
    }
    else{
        T = new TreeLinkNode(data);
        //构造左子树
        CreateBTree(T->left);
        //构造右子树
        CreateBTree(T->right);
    }
    return 0;
}
// 输出
void LevelOrder(TreeLinkNode *root){
    if(root == NULL){
        return;
    }//if
    TreeLinkNode *p = root,*q;
    while(p){
        q = p;
        // 横向输出
        while(q){
            cout<<q->val<<"->";
            q = q->next;
        }//while
        if(q == NULL){
            cout<<"NULL"<<endl;
        }//if
        p = p->left;
    }//while
}

int main() {
    Solution solution;
    TreeLinkNode* root(0);
    CreateBTree(root);
    solution.connect(root);
    LevelOrder(root);
}

[LeetCode]117.Populating Next Right Pointers in Each Node II

标签：层次遍历   leetcode   二叉树   

原文地址：http://blog.csdn.net/sunnyyoona/article/details/42127195