【题解】【BT】【Leetcode】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

思路：

这题看似Binary Tree Level Order Traversal，实则因为满二叉树的设定，相比要简单太多，提前计算好每层的节点数就OK

代码：

void connect(TreeLinkNode *root) {
if(root == NULL) return;//考虑自身为空
queue<TreeLinkNode*> nodeQ;
nodeQ.push(root);
int n = 1;
int i = 0;
while(!nodeQ.empty()){
TreeLinkNode* top = nodeQ.front();
nodeQ.pop();
if(++i != n){
top->next = nodeQ.front();
}else{
n *= 2;
i = 0;
}
if(top->left != NULL)//考虑左右子树为空（叶子节点）的情况
nodeQ.push(top->left);
if(top->right != NULL)
nodeQ.push(top->right);
}
}