Populating Next Right Pointers in Each Node I&&II ——II仍然需要认真看看
Populating Next Right Pointers in Each Node I
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
乍一看很难,理清思路后很简单的。
/**
* 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||root->left==NULL)
return;
root->left->next=root->right;
if(root->next!=NULL)
root->right->next=root->next->left;
connect(root->left);
connect(root->right);
return ;
}
};
II
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
/**
* 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(NULL == root) return;
TreeLinkNode* start;
TreeLinkNode* curNode;
TreeLinkNode* nextNode;
while(root != NULL){
start = findStartNodeNextLev(root);
curNode = start;
nextNode = findNextNodeNextLev(root, start);
while(nextNode != NULL){
curNode -> next = nextNode;
curNode = nextNode;
nextNode = findNextNodeNextLev(root, curNode);
}
root = start;
}
}
private:
TreeLinkNode* findNextNodeNextLev(TreeLinkNode* &cur, TreeLinkNode* curNextLev){
if(cur -> left == curNextLev && cur -> right != NULL){
return cur -> right;
}else{
while(cur -> next != NULL){
cur = cur -> next;
if(cur -> left != NULL && cur -> left != curNextLev) return cur -> left;
if(cur -> right != NULL && cur -> right != curNextLev) return cur -> right;
}
}
return NULL;
} TreeLinkNode* findStartNodeNextLev(TreeLinkNode* node){
if(NULL == node) return NULL;
if(node -> left != NULL) return node -> left;
return findNextNodeNextLev(node, node -> left);
}
};
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