[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

迭代

/**
* 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 *cur = root;
while (cur) {
TreeLinkNode *prev = NULL; //左兄弟
TreeLinkNode *next = NULL;//下一层的第一个点
while (cur) {
if (!next) {
next = cur->left?cur->left:cur->right;
}
if (cur->left) {
if (!prev) {
prev = cur->left;
}
prev->next = cur->left;
prev = cur->left;
}
if (cur->right) {
if (!prev) {
prev = cur->right;
}
prev->next = cur->right;
prev = cur->right;
}
cur = cur->next;
}
cur = next;
}
return ;
}
};

递归

/**
* 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 dummy(-1);
for (TreeLinkNode *cur = root, *prev = &dummy; cur; cur = cur->next) {
if (cur->left) {
prev->next = cur->left;
prev = cur->left;
}
if (cur->right) {
prev->next = cur->right;
prev = cur->right;
}
}
connect(dummy.next);
}
};