117. Populating Next Right Pointers in Each Node II

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;
        }
    }
}