Populating Next Right Pointers in Each Node

You are given a perfect binary tree where all leaves are on the same level, and every parent has two children. The binary tree has the following definition:

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *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.

Follow up:

• You may only use constant extra space.
• Recursive approach is fine, you may assume implicit stack space does not count as extra space for this problem.

Example 1:

Input: root = [1,2,3,4,5,6,7]
Output: [1,#,2,3,#,4,5,6,7,#]
Explanation: Given the above perfect binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.

Constraints:

• The number of nodes in the given tree is less than 4096.
• -1000 <= node.val <= 1000

A:

-------------method 1： layer wise , record nodes on each layer ------------------

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/
class Solution {
public:
    Node* connect(Node* root) {
        vector<Node*> layer;
        if(root)
            layer.push_back(root);
        while(!layer.empty())
        {
            vector<Node*> newLayer;
            for(int i =0;i<layer.size(); ++i)
            {
                if(i<layer.size()-1)
                    layer[i]->next = layer[i+1];
                if(layer[i]->left)
                {
                    newLayer.push_back(layer[i]->left);
                    newLayer.push_back(layer[i]->right);
                }
            }
            layer = newLayer;            
        }
        return root;
    }
};

------------------------------第二遍-------利用了已经建立了的next指针-----------

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/
class Solution {
public:
    Node* connect(Node* root) {
        // make use of the next node we just created
        if(!root)
            return NULL;
        Node* parent = root;
        Node* firstLeft = parent->left;
        while(firstLeft)
        {
            Node* upRunner=parent;
            Node* curRunner = firstLeft;
            while(curRunner)
            {
                curRunner->next = upRunner->right;
                curRunner = curRunner->next;
                if(upRunner->next)
                {
                    curRunner->next = upRunner->next->left;
                    upRunner = upRunner->next;
                }
                curRunner = curRunner->next;
            }            
            parent = firstLeft;
            firstLeft = firstLeft->left;
        }
        return root;
    }
};

Mistakes:
1: Big mistake:
     when we modify the structure of the tree, during recursive calling process, we are actually already changed them.
    i.e.  when we add the left subtree to the right child of the root,    we need first represent the right subtree, in order for further flatten.

2： 在set right 之后，要记得把left child改成null， 否则会leftchild 也指向一个指针。 你真SB

Learned:
1:  可以直接用Stack， 而不必用LinkedList 去模拟。