117. Populating Next Right Pointers in Each Node II

欢迎fork and star：Nowcoder-Repository-github

117. Populating Next Right Pointers in Each Node II

题目

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

解析

// Populating Next Right Pointers in Each Node
class Solution_117 {
public:
	//运行时间：8ms
    //占用内存：892k
	//使用层次遍历，每一层从左到右串接起来就行，每层最后一个元素next置NULL即可！
	void connect(TreeLinkNode *root) {
		if (!root)
		{
			return;
		}
		queue<TreeLinkNode*> que;
		que.push(root);

		while (!que.empty())
		{
			int size = que.size(); //每一层的大小

			TreeLinkNode* cur, *pre=NULL;
			while (size--)
			{
				cur= que.front();
				que.pop();

				if (pre)
				{
					pre->next = cur;		
				}
				pre = cur;
				if (cur->left)
				{
					que.push(cur->left);
				}
				if (cur->right)
				{
					que.push(cur->right);
				}
			}
			cur->next = NULL;

		}
	}

	void connect1(TreeLinkNode *root) {
		queue<TreeLinkNode*> q;
		if (!root)
			return;
		q.push(root);

		while (!q.empty()) {
			int size = q.size();
			for (int i = 0; i < size; i++) {
				TreeLinkNode* node = q.front();
				q.pop();
				if (i == size - 1)
					node->next = nullptr;
				else
					node->next = q.front();//当前节点已经出栈， q.front()为下一节点，避免记录上一次节点

				if (node->left)
					q.push(node->left);
				if (node->right)
					q.push(node->right);
			}
		}
	}
};

117. Populating Next Right Pointers in Each Node II