LeetCode OJ 116. Populating Next Right Pointers in Each Node

Posted 2020-08-26

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

 

 

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解答

利用已经连好的next指针进行广度优先遍历，注意是每次处理下一层，所以下一层为NULL的时候就不需要处理了。

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  struct TreeLinkNode *left, *right, *next;
 * };
 *
 */
void connect(struct TreeLinkNode *root) {
    struct TreeLinkNode *head, *next_head, *pNode, *pre_node;
    
    if(NULL == root){
        return;
    }
    root->next = NULL;
    next_head = root;
    while(1){
        head = next_head;
        next_head = next_head->left;
        if(NULL == next_head){
            break;
        }
        for(pNode = head; pNode != NULL; pNode = pNode->next){
            if(pNode != head){
                pre_node->next = pNode->left;
            }
            pNode->left->next = pNode->right;
            pre_node = pNode->right;
        }
        pre_node->next = NULL;
    }
}