[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 == -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);
}

【分析二】

对于一个左节点，相邻节点为父节点的右子节点。next指针指向父节点的右子节点。

对于一个当右节点稍微麻烦一些。相邻节点为父节点的右相邻节点的左子结点。父节点的右相邻节点还可能为空，所以需要判断一下。

// 父节点右相邻节点不为空
            if(next){
                connect(cur->right,next->left);
            }//if
            // 父节点右相邻节点为空
            else{
                connect(cur->right,NULL);
            }//else

next指针指向父节点的右相邻节点的左子结点或者空指针。

【代码二】

class Solution {
public:
    void connect(TreeLinkNode *root) {
        if(root == NULL){
            return;
        }//if
        connect(root,NULL);
    }//void
private:
    // cur 当前节点 next 右相邻节点
    void connect(TreeLinkNode *cur,TreeLinkNode *next) {
        if(cur == NULL){
            return;
        }//if
        else{
            cur->next = next;
        }//else
        // 左子树(连接的下一节点为父节点的右节点)
        if(cur->left){
            connect(cur->left,cur->right);
        }//if
        // 右子树(连接的下一节点为父节点相邻节点的左节点)
        if(cur->right){
            // 右相邻节点不为空
            if(next){
                connect(cur->right,next->left);
            }//if
            // 右相邻节点为空
            else{
                connect(cur->right,NULL);
            }//else
        }//if
    }//void
};