LeetCode 117:Populating Next Right Pointers in Each Node II

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

Subscribe to see which companies asked this question

//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);
	}
};

LeetCode 117:Populating Next Right Pointers in Each Node II_第1张图片

你可能感兴趣的:(LeetCode 117:Populating Next Right Pointers in Each Node II)