[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

解法一：

类似binary tree level traversal， 使用一个queue，将每一层的node存在queue中。具体见code。不过该方法不是constant space complexity。

class Solution {
public:
void connect(TreeLinkNode *root) {
if(!root) return;
queue<TreeLinkNode*> q;
q.push(root);

while(!q.empty()){
int len = q.size();
TreeLinkNode* cur=q.front(), *pre = q.front();
q.pop();
if(pre->left) q.push(pre->left);
if(pre->right) q.push(pre->right);
for(int i=1; i<=len-1; i++){
cur = q.front();
q.pop();
if(cur->left) q.push(cur->left);
if(cur->right) q.push(cur->right);
pre->next = cur;
pre = cur;
}
cur->next = NULL;
}
}
};

解法二：

recursion的思想。这里要想到root->right->next = root->next->left。这个也不是constant space。

/**
* 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;
if(root->left) root->left->next = root->right;
if(root->right) root->right->next = root->next?root->next->left:NULL;
connect(root->left);
connect(root->right);

}
};

解法三：

/**
* 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* parent = root;
if(!parent->left) return;

while(parent){
TreeLinkNode* levelHead = parent->left;
if(!levelHead) return;
TreeLinkNode* cur = levelHead;
TreeLinkNode* oldParent = parent;
while(cur){
if(cur==oldParent->left){
cur->next = parent->right;
cur = parent->right;
oldParent = parent;
parent = parent->next;
}else{
cur->next= parent?parent->left:NULL;
cur = cur->next;
oldParent = parent;
}
}
parent = levelHead;
}

}
};