【LeetCode】Populating Next Right Pointers in Each Node（I and II）

最新推荐文章于 2021-09-24 12:05:40 发布

转载最新推荐文章于 2021-09-24 12:05:40 发布 · 376 阅读

算法学习同时被 2 个专栏收录

104 篇文章

订阅专栏

【LeetCode】

100 篇文章

订阅专栏

本文探讨了如何在不增加额外空间的情况下填充二叉树节点的next指针，使其指向同一层的下一个节点。提供了两种情况下的解决方案：完美二叉树和一般二叉树，并附带详细的代码实现。

题目描述：

Populating Next Right Pointers in Each Node

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *next;
    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

You may only use constant extra space.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

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

这题目用DFS或者BFS的话都很简单，有趣的地方在于使用O（1）的空间复杂度解题。先贴一下I的DFS解法：

struct TreeLinkNode {
 int val;
 TreeLinkNode *left, *right, *next;
 TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
};
//recursion solution,DFS,isn't constant memory
class Solution {
public:
	void connect(TreeLinkNode *root) 
	{
		if (!root)
			return;
		if (root->left)
			conn(root->left, root->right);
	}
	void conn(TreeLinkNode *left, TreeLinkNode *right)
	{
		left->next = right;
		if (left->left)
		{
			conn(left->left, left->right);
			conn(left->right, right->left);
			conn(right->left, right->right);
		}
	}
};

要保持O（1）的空间复杂度，考虑用循环代替递归。第一个while循环深度，第二个while在一层中遍历。代码：

struct TreeLinkNode {
	int val;
	TreeLinkNode *left, *right, *next;
	TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
};
class Solution {
public:
	void connect(TreeLinkNode *root)
	{
		TreeLinkNode *leftWall = root;
		while (leftWall)
		{
			TreeLinkNode *across = leftWall;
			while (across)
			{
				if (across->left)
					across->left->next = across->right;
				if (across->right&&across->next)
					across->right->next = across->next->left;
				across = across->next;
			}
			leftWall = leftWall->left;
		}
	}
};

对题目II来说，在同层中遍历时保存下前一个节点，找到下一个存在的节点再进行赋值：

struct TreeLinkNode {
 int val;
 TreeLinkNode *left, *right, *next;
 TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
};

class Solution {
public:
	void connect(TreeLinkNode *root) 
	{
		TreeLinkNode *leftWall = root;
		while (leftWall)
		{
			TreeLinkNode *across = leftWall;
			TreeLinkNode *nextWall = NULL;
			TreeLinkNode *prev = NULL;
			while (across)
			{
				if (across->left&&across->right)
				{
					if (prev)
						prev->next = across->left;
					if (!nextWall)
						nextWall = across->left;
					across->left->next = across->right;
					prev = across->right;
				}
				else if (across->left)
				{
					if (prev)
						prev->next = across->left;
					if (!nextWall)
						nextWall = across->left;
					prev = across->left;
				}
				else if (across->right)
				{
					if (prev)
						prev->next = across->right;
					if (!nextWall)
						nextWall = across->right;
					prev = across->right;
				}
				across = across->next;
			}
			leftWall = nextWall;
		}
	}
};