Populating Next Right Pointers in Each Node II 解答

Question

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

思路与Populating Next Right Pointer 一样，仍是用四个指针 prevHead, prevCurrent, curHead, current层次遍历。

两层循环：

外层循环：traverse level by level

内层循环: traverse last level, link current level

/**
 * 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) {
        TreeLinkNode prevHead = root, prevCur = root;
        TreeLinkNode currentHead = null, current = null;
        while (prevHead != null) {
            prevCur = prevHead;
            // Find current head
            while (prevCur != null && prevCur.left == null && prevCur.right == null) {
                prevCur = prevCur.next;
            }
            if (prevCur == null) {
                break;
            } else if (prevCur.left != null) {
                currentHead = prevCur.left;
                current = currentHead;
                if (prevCur.right != null) {
                    current.next = prevCur.right;
                    current = current.next;
                }
            } else {
                currentHead = prevCur.right;
                current = currentHead;
            }
            // link current level
            prevCur = prevCur.next;
            while (prevCur != null) {
                if (prevCur.left != null) {
                    current.next = prevCur.left;
                    current = current.next;
                }
                if (prevCur.right != null) {
                    current.next = prevCur.right;
                    current = current.next;
                }
                prevCur = prevCur.next;
            }
            // reset
            prevHead = currentHead;
        }
        
    }
}