Leetcode Populating Next Right Pointers in Each Node II

最新推荐文章于 2020-11-14 09:32:10 发布

最新推荐文章于 2020-11-14 09:32:10 发布 · 197 阅读

本文提供了一种解决LeetCode上填充每个节点的下一个右侧节点指针 II问题的方法，该问题是对填充每个节点的下一个右侧节点指针问题的扩展，适用于任意二叉树。文中给出了两种简洁的实现方式，均只使用常数额外空间。

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,

After calling your function, the tree should look like:

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

如何处理各个细节的问题很重要，决定了能否写出更加简洁的程序来。

void connect(TreeLinkNode *root) 
	{
		while (root && !root->left && !root->right) root = root->next;
		if (root == NULL) return;
		
		TreeLinkNode *rightSibling;
		TreeLinkNode *p1 = root;
		while (p1) 
		{
			TreeLinkNode *p_next = p1->next;
			while (p_next && !p_next->left && !p_next->right) 
				p_next = p_next->next;

			if (p_next)
				rightSibling = p_next->left? p_next->left:p_next->right;
			else rightSibling = NULL;

			if (p1->left && p1->right)
			{
				p1->left->next = p1->right;
				p1->right->next = rightSibling;
			}
			else if (p1->right)
			{
				p1->right->next = rightSibling;
			}
			else
			{
				p1->left->next = rightSibling;
			}

			p1 = p_next;
		}
		if (root->left)
			connect(root->left);
		else if(root->right)
			connect(root->right);
	}

Leetcode论坛上的一个简洁的程序，写得太好了，编程功底就在微小的差别中体现出来了。

具体思想是和我上面的程序一样的。

//LeetCode论坛上的
	// the link of level(i) is the queue of level(i+1)
	void connect2(TreeLinkNode * n) {
		while (n) {
			TreeLinkNode * next = NULL; // the first node of next level
			TreeLinkNode * prev = NULL; // previous node on the same level
			for (; n; n=n->next) {
				if (!next) next = n->left?n->left:n->right;

				if (n->left) {
					if (prev) prev->next = n->left;
					prev = n->left;
				}
				if (n->right) {
					if (prev) prev->next = n->right;
					prev = n->right;
				}
			}
			n = next; // turn to next level
		}
	}

//2014-2-17 update
	void connect(TreeLinkNode *root) 
	{
		while (root)
		{
			TreeLinkNode *next = NULL;
			TreeLinkNode *pre = NULL;
			for ( ; root; root = root->next)
			{
				helper(root->left, next, pre);
				helper(root->right, next, pre);
			}
			root = next;
		}
	}
	void helper(TreeLinkNode *&node, TreeLinkNode *&n, TreeLinkNode *&p)
	{
		if (!node) return;
		if (!n) n = node;
		if (p) p->next = node;
		p = node;
	}