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LeetCode 117:Populating Next Right Pointers in Each Node II

时间:2015-12-31 01:47:10      阅读:203      评论:0      收藏:0      [点我收藏+]

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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

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//二叉树进行层次遍历
//判断最底层的条件:如果当前访问的层有任一节点存在子节点,说明当前层不是最底层。
class Solution{
public:
	TreeLinkNode* nextLeft;
	bool hasToTheEnd;
	void connect(TreeLinkNode *root)
	{
		if (!root)  return;

		hasToTheEnd = false;
		int level = 1;
		while (!hasToTheEnd)
		{
			hasToTheEnd= true;
			nextLeft= NULL;
			VisitLevel(root, level);
			++level;
		}
	}

	void VisitLevel(TreeLinkNode* node, int level)
	{
		if (level == 1)
		{
			if (nextLeft != NULL)
			{
				nextLeft->next = node;
			}
			nextLeft = node;

			if (node->left != NULL || node->right != NULL)
			{
				hasToTheEnd= false;
			}
		}

		if (node->left)	  VisitLevel(node->left, level - 1);
		if (node->right)  VisitLevel(node->right, level - 1);
	}
};

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LeetCode 117:Populating Next Right Pointers in Each Node II

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原文地址:http://blog.csdn.net/geekmanong/article/details/50439036

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