LeetCode116 Populating Next Right Pointers in Each Node 二叉树下一指针添加-优快云博客

本文链接：https://blog.youkuaiyun.com/m0_37518259/article/details/81202261

本文介绍了一种在完美二叉树中填充每个节点的next指针的方法，使得每个节点指向其右侧相邻节点。提供了两种解决方案：一种是利用队列进行层次遍历，另一种是采用递归方式。这两种方法均能满足题目要求，即仅使用常数额外空间。

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.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

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

题源：here；完整实现：here

思路：

两种方案：1 使用队列；2 递归

1 使用队列

同第102题，我们遍历二叉树的每一层，并将该层中的各个结点连接起来。

void connect(TreeLinkNode *root) {
	if (!root) return;
	queue<TreeLinkNode *> q;
	q.push(root);
	TreeLinkNode *last = root;
	while (q.size()){
		TreeLinkNode* curr = q.front();
		q.pop();
		if(curr->left) q.push(curr->left);
		if(curr->right) q.push(curr->right);
		if (last == curr){
			if (q.size()) last = q.back();
		}
		else{
			curr->next = q.front();
		}
	}
}

2 递归

这个递归很有意思，为了达到同层相连的效果需要使用到root->next来搭桥。

void connect2(TreeLinkNode *root){
	if (!root) return;
	if (root->left) root->left->next = root->right;
	if (root->right) root->right->next = root->next ? root->next->left : NULL;
	connect(root->left);
	connect(root->right);
}