leetcode_question_117 Populating Next Right Pointers in Each Node II

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#leetcode #Populating Next Righ #bfs

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本文探讨了如何在不使用额外空间的情况下填充二叉树节点的next指针，使其指向同一层级的下一个节点。提供了两种解决方案：一种是使用广度优先搜索（BFS），另一种是递归方法。

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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

leetcode_question_116 Populating Next Right Pointers in Each Node的代码一点都没改在这道题就ACCEPTED，我感觉好失败啊，所以，上一题应该有更简单的方法。

BFS:

/**
 * Definition for binary tree with next pointer.
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root==NULL)return;
    	vector<TreeLinkNode*> vec;
		vec.push_back(root);
		int count = 1;
		while(!vec.empty())
		{
			if(count > 1) vec[0]->next = vec[1];
			else vec[0]->next = NULL;
			if(vec[0]->left) vec.push_back(vec[0]->left);
			if(vec[0]->right) vec.push_back(vec[0]->right);
			vec.erase(vec.begin());
			count--;

			if(count == 0)
				count = vec.size();
		}
    }
};

Recursive:

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    TreeLinkNode* rightNeighbor(TreeLinkNode* left)
{
	if(left->next == NULL) return NULL;
	else if(left->next->left)
		return left->next->left;
	else if(left->next->right)
		return left->next->right;
	else return rightNeighbor(left->next);
}

void connectHelper(TreeLinkNode* root)
{
	if(root)
	{
		if(root->right)
			root->right->next = rightNeighbor(root);
		if(root->left)
		{
			if(root->right)
				root->left->next = root->right;
			else
				root->left->next = rightNeighbor(root);
		}
		connectHelper(root->right);
		connectHelper(root->left);
	}
}

void connect(TreeLinkNode *root) {
        if(root)
		{
			root->next = NULL;
			if(root->right)
				root->right->next = NULL;
			if(root->left)
			{
				if(root->right) 
					root->left->next = root->right;
				else
					root->left->next = NULL;
			}
			connectHelper(root->right);
			connectHelper(root->left);
		}
    }
};