# 【leetcode】Populating Next Right Pointers in Each Node

Question ：

Given a binary tree

```    struct TreeLinkNode {
}```

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```

Anwser 1：    Travesal

```/**
* Definition for binary tree with next pointer.
*  int val;
*  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(NULL == root){
return;
}

while(left && left->left && left->right)
{
root = left;
while(root)
{
root->left->next = root->right;     // connect two nodes of root
if(root->next){
root->right->next = root->next->left;   // connect two isolated nodes
}

root=root->next;    // traversal the same level line
}
left=left->left;
}
}
};```

Anwser 2：    Recursive

```/**
* Definition for binary tree with next pointer.
*  int val;
*  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(NULL == 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);
}
};```

Anwser 3： Queue

```/**
* Definition for binary tree with next pointer.
*  int val;
*  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(NULL == root){
return;
}

if(root != NULL){
Q.push(root);
}

int row = 1;
int count = 0;
while(!Q.empty()){
Q.pop();
count++;

if(tmp->left) Q.push(tmp->left);
if(tmp->right) Q.push(tmp->right);

if(count == row){
tmp->next = NULL;
count = 0;
row *= 2;
} else {
tmp->next = Q.front();
}
}
}
};```

1） Queue > Traversal > Recursive

2） Queue 没有递归的层次限制，可以使用很大的二叉树

0 条评论

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