116. Populating Next Right Pointers in Each Node

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

递归
先连接左右子节点，然后连接右子节点和右节点(如果有的话)的左子节点
接下来递归下一层即可

class Solution {
public:
    void connect(TreeLinkNode *root) {
        if (root == NULL || root->left == NULL) //We may assume that it is a perfect binary tree,so (root->left == NULL) equal (root->left == NULL && root->right == NULL)
            return;

        root->left->next = root->right;

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

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

迭代

class Solution {
public:
    void connect(TreeLinkNode *root) {
        if(root == NULL || root->left == NULL)
            return;
            
        TreeLinkNode* temp = root;
        while(temp->left){
            //每次处理一层
            while(temp){
                temp->left->next = temp->right;
                if(temp->next)
                    temp->right->next= temp->next->left;
                temp = temp->next;  
            }
            //root记录所在层
            root = root->left;
            temp = root;
        }
    }
};