12
08/2015
[LeetCode] Populating Next Right Pointers in Each Node
Populating Next Right Pointers in Each Node
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
解题思路：
注意这里说的是完美二叉树，即满二叉树。分层来做。记录上一层的第一个节点，然后遍历上一层节点U和下一层节点L，将U的左孩子节点指向U的右孩子节点，L的右孩子节点指向U的下一个节点（如果存在的话）的左孩子节点。
/**
* 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* upLayerFirstNode = root;
while(upLayerFirstNode->left!=NULL){
TreeLinkNode* upLayerNode = upLayerFirstNode;
TreeLinkNode* node = upLayerFirstNode->left;
while(true){
node->next = upLayerNode->right;
node = node->next;
upLayerNode = upLayerNode->next;
if(upLayerNode==NULL){
break;
}
node->next = upLayerNode->left;
node = node->next;
}
upLayerFirstNode = upLayerFirstNode->left;
}
}
};
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