Populating Next Right Pointers in Each Node II

标签：leetcode   算法   

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

点击打开原题链接

基本上是层次遍历稍微修改下就可以了，算法比较简单，注意下边界条件即可，也可以用把每一层放入vector<vector<> >中去，但是会开辟新的内存，这里

直接判断是否在一层上，代码稍微复杂一点点。

struct TreeLinkNode 
 {
	  int val;
	  TreeLinkNode *left, *right, *next;
	  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
	};
 struct node
 {
	 int level;
	 TreeLinkNode* TLN;
 };
	
	class Solution 
	{
	public:
		void connect(TreeLinkNode *root) 
		{
			if (root == NULL)
			{
				return;
			}
			deque<node> deque_TreeLinkNode;
			node pre,current;
			pre.level = 0;
			pre.TLN = root;
			deque_TreeLinkNode.push_back(pre);
			pre.TLN = NULL;
			while(!deque_TreeLinkNode.empty())
			{
				node temp = deque_TreeLinkNode.front();
				
				if (temp.TLN->left != NULL)
				{
					node left;
					left.level = temp.level+1;
					left.TLN = temp.TLN->left;
					deque_TreeLinkNode.push_back(left);
				}
				if (temp.TLN->right != NULL)
				{
					node right;
					right.level = temp.level+1;
					right.TLN = temp.TLN->right;
					deque_TreeLinkNode.push_back(right);
				}

				current= deque_TreeLinkNode.front();
				deque_TreeLinkNode.pop_front();
				if (pre.TLN == NULL)
				{
					pre.TLN = current.TLN;
					pre.level = current.level;
				//	deque_TreeLinkNode.pop_front();
					continue;
				}
				else
				{
					if (current.level == pre.level)
					{
						
						pre.TLN->next = current.TLN;
						pre=current;
						continue;
					//	current->TLN = NULL;
					//	current->level = 0;
					}
					else
					{
						pre.TLN->next = NULL;
						pre = current;
					}

				}

			}
			pre.TLN->next = NULL;
			

		}
	private:
		
	};

Populating Next Right Pointers in Each Node II

标签：leetcode   算法   

原文地址：http://blog.csdn.net/hongkangwl/article/details/39807507