Populating Next Right Pointers in Each Node II

Total Accepted: 57390 Total Submissions: 175596 Difficulty: Hard
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

思路

• 来自于I
• 稍微改变一下,找到右边下一层的最左点

/**
 * 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) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if (root == null) {
            return;
        }

        TreeLinkNode p = root.next;

        while (p != null) {
            if (p.left != null) {
                p = p.left;
                break;
            }
            if (p.right != null) {
                p = p.right;
                break;
            }
            p = p.next;
        }

        if (root.right != null) {
            root.right.next = p;
        }

        if (root.left != null) {
            root.left.next = root.right == null ? p : root.right;
        }

        connect(root.right);
        connect(root.left);
    }
}