leetcode Populating Next Right Pointers in Each Node

看图就知道想要做什么事了。

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *next;
    }

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

初始所有人的next都是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).

要求常数额外空间。

 

一个递归就搞定了，就是递归让每一个节点他的左右子树通过next链接，直至到最后一层，然后递归左右节点，继续让他们的左右子树通过next链接。

/**
 * 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:

//每一层把left的right的next设置为right的left
void lr2rl(TreeLinkNode *root)
{
    if (!root) return;
    TreeLinkNode *lr = root -> left, *rl = root -> right;
    
    while(lr && rl)
    {
        lr -> next = rl;
        lr = lr -> right;
        rl = rl -> left;
    }
    lr2rl(root -> left);
    lr2rl(root -> right);
}
    void connect(TreeLinkNode *root) 
    {
        lr2rl(root);
    }
};

 也可以每层每层的完成：

/**
 * 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) return ;
        TreeLinkNode *lf = root;
        while (root)
        {
            if (root -> left)
                root -> left -> next = root -> right;
            if (root -> next && root -> next -> left)
                root -> right -> next = root -> next -> left;
            root = root -> next;
        }
        connect(lf -> left);
        connect(lf -> right);
    }
};

 可以用非递归的方法，把每一行当做一个queue，每次从最左边开始处理到最右边。记录下一层的最左边。继续。直至null

/**
 * 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:
    //非递归法，将做好的每一行当做一个queue
    void connect(TreeLinkNode *root) 
    {
        if (!root) return ;
        TreeLinkNode *lf;
        while(root)
        {
            lf = root -> left;
            while(root)
            {
                if (lf)
                    root -> left -> next = root -> right;
                if (root -> next && root -> next -> left)
                    root -> right -> next = root -> next -> left;
                root = root -> next;
            }
            root = lf;
        }
    }
};