Populating Next Right Pointers in Each Node II

本文介绍了一种在任意二叉树中填充每个节点的Next指针的方法,确保相同层级的相邻节点能够通过Next指针相连。该方法只使用常数额外空间,并遵循先右后左的递归策略。

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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


/**
 * 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 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) {
			if (root.right != null) {
				root.left.next = root.right;
			} else {
				root.left.next = p;
			}
		}
		connect(root.right);
		connect(root.left);
	}
}
 
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