《算法分析与设计》Week 11

116. Populating Next Right Pointers in Each Node

Description:

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

Solution:

一、题意理解

     给定一颗完全二叉树，将每一层的节点的 *next指针都指向同一层右面的节点，最右面一个节点的*next指向NULL

二、分析

     1、可以递归地来将父节点的两个孩子连接起来，然后将父节点的右孩子和父节点的右兄弟节点的左孩子连接起来，如果父节点没有右兄弟节点，则其右孩子*next = NULL

     2、递归比较麻烦而且在时间和空间上都不尽人意，像这种一层一层的，明显可以用到非递归的层序遍历，来依次连接同一层的节点。

     3、代码如下：

/**
 * 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) {
        queue<TreeLinkNode *> q;
        TreeLinkNode *tmp;
        int level;
        
        if(root != NULL)
            q.push(root);
        while(!q.empty())
        {
            level = q.size();
            while(level--)
            {
                tmp = q.front();
                q.pop();
                if(level == 0)
                    tmp->next = NULL;
                else
                    tmp->next = q.front();
                if(tmp->left)
                    q.push(tmp->left);
                if(tmp->right)
                    q.push(tmp->right);
            }
        }
    }
};