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(root == NULL) return;
TreeLinkNode *p = root -> next;
while(p){
if(p -> left){
p = p -> left;
break;
}
if(p -> right){
p = p -> right;
break;
}
p = p -> next;
}
if(root -> right)
root ->right -> next = p;
if(root -> left)
root -> left -> next = root -> right ? root -> right : p;
connect(root -> right);
connect(root -> left);
}
};
本文讨论了如何在不使用额外空间的情况下,为任意给定的二叉树节点添加下级链接,以实现树状结构的遍历。通过迭代方法从根节点开始,逐步为每个节点设置指向其子节点的链接。
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