Q117. Populating Next Right Pointers in Each Node II
直达:https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/description/
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
分析
对于任意一棵树来说,可以分成下面三种情况
若左子树存在,右子树也存在,左子树的next是右子树,右子树的next是父节点的兄弟节点的第一个孩子(root的大侄子);
若左子树存在,右子树不存在,左子树的next是父节点的兄弟节点的第一个孩子;
若左子树不存在,右子树存在,右子树的next是父节点的兄弟节点的第一个孩子;
所以问题的关键是寻找root的大侄子,依次查看root->next是否存在左右子树,否则,root = root->next;
同样,可以通过递归操作完成。
/**
* 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 *temp = root->next;
while(temp){
if(temp->left){
temp = temp->left;
break;
}else if(temp->right){
temp = temp->right;
break;
}
temp = temp->next;
}
if(root->right) root->right->next = temp;
if(root->left) root->left->next = (root->right)?(root->right):(temp);
connect(root->right);
connect(root->left);
}
};
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