LeetCode 116:Populating Next Right Pointers in Each Node

本文链接：https://blog.csdn.net/geekmanong/article/details/50437976

本文介绍了一种使用常数额外空间在满二叉树中填充每个节点的下一个指针的方法，使得所有节点的下一个指针都指向其右侧的节点。详细解释了算法的实现过程，并通过示例展示了遍历后的树结构。

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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

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//对二叉树进行先序遍历，根据题述，给定的是满二叉树，左孩子为空，则右孩子一定为空，所以左孩子为空，则return
//如果左孩子不为空，则右孩子一定不为空，所以链接左孩子和右孩子即可（左孩子的next赋值为右孩子）;
//由于先序遍历，所以父节点的next比子节点的next先被设置，故父节点不同的两个子节点进行连接，就可以用到父节点的next，
//即root->right->next = root->next->left;
class Solution {
public:
	void connect(TreeLinkNode *root) {
		if (root == NULL || root->left==NULL)
			return;

		if (root->left != NULL)
			root->left->next = root->right;

		if (root->right != NULL && root->next != NULL)
			root->right->next = root->next->left;
		else if (root->right != NULL)
			root->right->next = NULL;

		connect(root->left);
		connect(root->right);
	}
};