【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 = ;
     int i = ;
     while(!nodeQ.empty()){
         TreeLinkNode* top = nodeQ.front();
         nodeQ.pop();
         if(++i != n){
             top->next = nodeQ.front();
         }else{
             n *= ;
             i = ;
         }
         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);
     }
 }