LeetCode刷题笔录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

是前一道题的延伸，这里树不再是complete的了，而可以是任意形式的。

需要做些改动。在找下一层的head pointer时要判断current有哪个孩子；而且也不一定是记录前一个node的right child了，而是记录最近的一个没有更新其next的node。

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        if(root == null)
            return;
        TreeLinkNode nextLineHead = root;
        
        while(nextLineHead != null){
            TreeLinkNode previousTail = null, current = nextLineHead;
            nextLineHead = null;
            
            while(current != null){
                //update the next line header if the current node has a child
                if(nextLineHead == null){
                    if(current.left != null)
                        nextLineHead = current.left;
                    else if(current.right != null)
                        nextLineHead = current.right;
                }
                
                //if the current node has two children, link the left to the right
                if(current.left != null && current.right != null)
                    current.left.next = current.right;
                
                //link the previousTail node to one of current node's children if possible
                if(previousTail != null){
                    if(current.left != null && current.right != null){
                        previousTail.next = current.left;
                        previousTail = current.right;
                    }
                    else if(current.left != null){
                        previousTail.next = current.left;
                        previousTail = current.left;
                    }
                    else if(current.right != null){
                        previousTail.next = current.right;
                        previousTail = current.right;
                    }
                }
                else if(nextLineHead != null){
                    if(nextLineHead.next != null)
                        previousTail = nextLineHead.next;
                    else
                        previousTail = nextLineHead;
                }
                //move to the next node
                current = current.next;
            }
        }
    }
    
}