Question :
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:
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.
* 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(NULL == root){
return;
}
TreeLinkNode *left = root;
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.
* 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) {
// 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.
* 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) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(NULL == root){
return;
}
queue<TreeLinkNode *> Q;
if(root != NULL){
Q.push(root);
}
int row = 1;
int count = 0;
while(!Q.empty()){
TreeLinkNode *tmp = Q.front();
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();
}
}
}
};
注意点:
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