LeetCode 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

思路：

本题树为 完美树， 每次保存每一层的第一个节点， 执行到下一层时，上一层已经连接好了。   使用递归的话，代码更少。

非递归:

/**
* 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 == NULL)
return;
stack<TreeLinkNode *> MyStack;
MyStack.push(root);
while (!MyStack.empty()) {
TreeLinkNode *head = MyStack.top();
MyStack.pop();

if (head->left != NULL)
MyStack.push(head->left);

while (head != NULL) {
if (head->left == NULL)
break;
head->left->next = head->right;
if (head->next != NULL)
head->right->next = head->next->left;
head = head->next;
}
}
}
};

递归：

class Solution {
public:
void connect(TreeLinkNode *root) {
if (root == NULL)
return;

if (root->left != NULL) {
root->left->next = root->right;
root->right->next = root->next ? root->next->left : NULL;
}

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