Follow up for problem "Populating Next Right Pointers in Each Node".
What if the given tree could be any binary tree? Would your previous solution still work?
Note:
You may only use constant extra space.
For example,
Given the following binary tree,

         1
       /  \
      2    3
     / \    \
    4   5    7
After calling your function, the tree should look like:
         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL

Solution：Iterative 每层遍历

思路：Iterative 每层遍历，再遍历同层上各子树block，并keep一个prev用来连接

屏幕快照 2017-09-06 下午6.56.58.png

Time Complexity: O(N) Space Complexity: O(1)

Solution Code：

public class Solution {
    //based on level order traversal
    public void connect(TreeLinkNode root) {

        TreeLinkNode head = null; //head of the next level
        TreeLinkNode prev = null; //the leading node on the next level
        TreeLinkNode cur = root;  //current node of current level

        while (cur != null) { // iterate on next level  
            while (cur != null) { //iterate on the current level, different blocks
                //left child
                if (cur.left != null) {
                    if (prev != null) {
                        prev.next = cur.left;
                    } else {
                        head = cur.left;
                    }
                    prev = cur.left;
                }
                //right child
                if (cur.right != null) {
                    if (prev != null) {
                        prev.next = cur.right;
                    } else {
                        head = cur.right;
                    }
                    prev = cur.right;
                }
                //move to next node
                cur = cur.next;
            }
            
            //move to next level
            cur = head;
            head = null;
            prev = null;
        }
    }
}

Solution Code Round1

public class Solution {
    public void connect(TreeLinkNode root) {
        
        TreeLinkNode next_l_head = null;
        TreeLinkNode prev = null;
        TreeLinkNode post = null;
        TreeLinkNode cur = root;
        
        while(cur != null) { // different level
            while(cur != null) { // different nodes on same level
                // left child
                if(cur.left != null) {
                    post = cur.left;
                    if(prev != null) prev.next = post;
                    else next_l_head = post;
                    prev = post;
                }
                // right child
                if(cur.right != null) {
                    post = cur.right;
                    if(prev != null) prev.next = post;
                    else next_l_head = post;
                    prev = post;
                }
                // move to next node
                cur = cur.next;
            }
            // move to next level
            cur = next_l_head;
            next_l_head = null;
            prev = null;
        }
    }
}