LeetCode 116: 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,

    1   /  \  2    3 / \  / \4  5  6  7

After calling your function, the tree should look like:

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

解题思路

输入是一个满二叉树（perfect binary tree）。

思路一：利用一个队列（Queue）对二叉树进行层次遍历，在遍历过程中将同一层的节点通过 next 之间链接起来。但是所需空间不是 constant extra space。

思路二：递归进行深度优先搜索。但是递归是需要额外的栈空间的，所需空间不是 constant extra space。代码如下：

/** * 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) {        if (root == NULL) return;        if (root->left != NULL) {            root->left->next = root->right;        }        if(root->right != NULL && root->next != NULL) {            root->right->next = root->next->left;        }        connect(root->left);        connect(root->right);    }};

思路三：借用 next 指针，做到不需要队列就能完成广度优先搜索（BFS）：如果当前层所有结点的 next 指针已经设置好了，那么据此，下一层所有结点的 next 指针也可以依次被设置。代码如下：

/** * 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) {        while (root != NULL) {            if (root->left == NULL) return;            TreeLinkNode *p = root;            while (p != NULL) {                p->left->next = p->right;                if (p->next != NULL) {                    p->right->next = p->next->left;                }                p = p->next;            }            root = root->left;        }     }};