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:一层一层的连接,同一个父节点的左子树的邻居是其父节点的右子树 父节点右子树的邻居是 父节点邻居的左子树
我用的递归
void connect(TreeLinkNode *root) {
if(root == NULL) return;
TreeLinkNode * parent = root;
TreeLinkNode * cur = parent->left;
while(parent != NULL && cur != NULL)
{
cur = cur->next = parent->right; //左子树的邻居是同一个parent的右子树
cur->next = (parent->next == NULL) ? NULL : parent->next->left; //右子树的邻居 是parent的邻居的左子树
parent = parent->next;
cur = (parent == NULL) ? NULL : parent->left;
}
connect(root->left); //连接下一层
}
大神的非递归代码:外面加一圈对层循环
void connect(TreeLinkNode *root) {
if(!root)
return;
while(root -> left)
{
TreeLinkNode *p = root;
while(p)
{
p -> left -> next = p -> right;
if(p -> next)
p -> right -> next = p -> next -> left;
p = p -> next;
}
root = root -> left;
}
}
Follow up for problem "Populating Next Right Pointers in Each Node".
What if the given tree could be any binary tree? Would your previous solution still work?
Note:
- You may only use constant extra space.
For example,
Given the following binary tree,
1
/ \
2 3
/ \ \
4 5 7
After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ \
4-> 5 -> 7 -> NULL
思路:非完全二叉树。关键是每一层要判断邻居是哪一个,每层的第一个有数字的位置也要定位。
我的代码,根据上一题的思路,用的非递归。对父节点左右子树都空,左子树空,右子树空,都非空分类处理。
void connect2(TreeLinkNode *root) {
if(root == NULL) return;
TreeLinkNode * newroot = root;
while(newroot != NULL) //对层循环
{
TreeLinkNode * p = newroot; //当前层父节点推进
TreeLinkNode * cur = NULL; //当前连接层当前指针位置
newroot = NULL; //下一层第一个父节点的位置
while(p != NULL)
{
if(p->left == NULL && p->right == NULL);
else if(p->left == NULL)
{
if(cur == NULL)
newroot = cur = p->right;
else
cur = cur->next = p->right;
}
else if(p->right == NULL)
{
if(cur == NULL)
newroot = cur = p->left;
else
cur = cur->next = p->left;
}
else
{
if(cur == NULL)
newroot = cur = p->left;
else
cur = cur->next = p->left; cur = cur->next = p->right;
}
p = p->next;
}
}
}
大神的代码,处理的时候只要分左子树是否空,和右子树是否空即可。不需要分那么多情况。
public class Solution {
public void connect(TreeLinkNode root) { while(root != null){
TreeLinkNode tempChild = new TreeLinkNode(); //该层的伪头结点,方便定位下一层第一个值的位置
TreeLinkNode currentChild = tempChild; //这两个值不用每次分配,在外面分配,内部循环使用即可
while(root!=null){ //只分两种情况就行了
if(root.left != null) { currentChild.next = root.left; currentChild = currentChild.next;}
if(root.right != null) { currentChild.next = root.right; currentChild = currentChild.next;}
root = root.next;
}
root = tempChild.next;
}
}
}