Leetcode:populating_next_right_pointers_in_each_node题解

一、     题目

对于结构体:struct TreeLinkNode {

TreeLinkNode *left;

TreeLinkNode *right;

TreeLinkNode *next;

}

填充全部的结点假设存在右側的节点。则指上(next)右节点。假设没有右側的节点。那么就指上NULL，最初都指上NULL。

提示:仅仅能使用给定的空间，而且你能够觉得给定的二叉树是一个完美的二叉树。

比如：

1

/ \

2   3

/ \   \

4  5    7

处理后:

1 -> NULL

/ \

2 -> 3 -> NULL

/ \   \

4-> 5 -> 7 -> NULL

二、     分析

1、         空节点就直接返回

2、         左节点非空则连接右节点

3、         子节点连接兄弟节点的子节点。则root.right.next = root.next.left;

/**
 * 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;
        if(root->left!=NULL) root->left->next=root->right;
        if(root->right!=NULL&&root->next!=NULL)
        	root->right->next=root->next->left;

        connect(root->left);
        connect(root->right);
    }
};

BFS
/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        //breadth-first traversal
        queue<TreeLinkNode *> bfsq;
        int levelCnt=0;
        int level2=0;

        if(root==NULL) return;
        bfsq.push(root);
        levelCnt++;
        TreeLinkNode *prevList=NULL;
        TreeLinkNode *topS=NULL;

        while(!bfsq.empty())
        {
            topS=bfsq.front();
            bfsq.pop();
            levelCnt--;

            if(topS->left!=NULL && topS->right!=NULL)
            {
                bfsq.push(topS->left);
                level2++;
                bfsq.push(topS->right);
                level2++;
            }

            if(prevList!=NULL)
                prevList->next=topS;
            prevList=topS;

            if(levelCnt==0)
            {
                levelCnt=level2;
                level2=0;
                prevList=NULL;
            }

        }
    }
};