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

一个完美的二叉树，那就不用考虑很多东西。只要pre->left存在，就说明还有下一层，就可以一直循环下去。时间复杂度O(n)，空间复杂度O(1)

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

117. Populating Next Right Pointers in Each NodeII

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,

         1       /  \      2    3     / \    \    4   5    7

After calling your function, the tree should look like:

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

跟上一题意思是一样的，只是改了一些条件，这次不再是一颗完美的二叉树了。

记录行开始的结点pre，记录上一行的行进结点head，记录当前行的行进结点cur。时间复杂度O(n)，空间复杂度O(1)

/** * 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) return;        TreeLinkNode* head = root;        TreeLinkNode* pre = NULL;        TreeLinkNode* cur = NULL;        while (head) {            pre = new TreeLinkNode(0);            cur = pre;            while (head) {                if (head->left) {                    cur->next = head->left;                    cur = cur->next;                }                if (head->right) {                    cur->next = head->right;                    cur = cur->next;                }                head = head->next;            }            head = pre->next;        }        return;    }};