LeetCode——Populating Next Right Pointers in Each Node

标签：leetcode

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *next;
    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

         1
       /        2    3
     / \  /     4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL
       /        2 -> 3 -> NULL
     / \  /     4->5->6->7 -> NULL

题目： 给定一个二叉树(假设是完全二叉树)，把每个节点的next指针指向其右侧节点。

思路：首先想到的是，层序遍历树，在遍历的同时添加节点对右侧节点的指针。

另一种简洁的方法是采用递归来实现，间单直观。

	public void connect(TreeLinkNode root) {
		if (root == null)
			return;
		if (root.left != null)
			root.left.next = root.right;
		if (root.right != null)
			root.right.next = root.next == null ? null : root.next.left;
		connect(root.left);
		connect(root.right);
	}

	// Definition for binary tree with next pointer.
	public class TreeLinkNode {
		int val;
		TreeLinkNode left, right, next;

		TreeLinkNode(int x) {
			val = x;
		}
	}

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LeetCode——Populating Next Right Pointers in Each Node

标签：leetcode

原文地址：http://blog.csdn.net/laozhaokun/article/details/38357015