leetcode之Populating Next Right Pointers in Each Node

题目大意:

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

意思就是:
给定一棵满二叉树,要使在一个平面上的每个节点的next指针指向该节点的右方的节点,如果没有,则指向NULL.

解体思路:

由于是满二叉树, 所以对于每个节点, 如果该节点是父亲的左孩子,则它的next指向父亲的右子树; 如果该节点是父亲的右孩子,则它的next指向该节点父亲的next的左孩子.

代码如下:

/**
* 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;
}
root->next = NULL;
queue<TreeLinkNode*> storage;
storage.push(root);
while(!storage.empty()){
TreeLinkNode* current = storage.front();
storage.pop();
if(current->left){
current->left->next = current->right;
}
if(current->next && current->right){
current->right->next = current->next->left;
}
else if(current->right){
current->right->next = NULL;
}
if(current->left){
storage.push(current->left);
}
if(current->right){
storage.push(current->right);
}
}
}
};