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社区首页 >专栏 >LeetCode 0117 - Populating Next Right Pointers in Each Node II

LeetCode 0117 - Populating Next Right Pointers in Each Node II

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Reck Zhang
发布2021-08-11 10:54:18
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发布2021-08-11 10:54:18
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文章被收录于专栏:Reck Zhang

Populating Next Right Pointers in Each Node II

Desicription

Follow up for problem “Populating Next Right Pointers in Each Node“.

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

  • You may only use constant extra space.

For example, Given the following binary tree,

代码语言:javascript
复制
     1
   /  \
  2    3
 / \    \
4   5    7

After calling your function, the tree should look like:

代码语言:javascript
复制
     1 -> NULL
   /  \
  2 -> 3 -> NULL
 / \    \
4-> 5 -> 7 -> NULL

Solution

代码语言:javascript
复制
/**
 * 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) {
        TreeLinkNode* cur = root;
        TreeLinkNode* nextLevelHead = NULL;
        TreeLinkNode* nextLevelPre = NULL;
        while(cur != NULL) {
            while(cur != NULL) {
                if(cur->left != NULL) {
                    if(nextLevelPre != NULL)
                        nextLevelPre->next = cur->left;
                    else
                        nextLevelHead = cur->left;
                    nextLevelPre = cur->left;
                }
                if(cur->right != NULL) {
                    if(nextLevelPre != NULL)
                        nextLevelPre->next = cur->right;
                    else
                        nextLevelHead = cur->right;
                    nextLevelPre = cur->right;
                }
                cur = cur->next;
            }
            cur = nextLevelHead;
            nextLevelHead = NULL;
            nextLevelPre = NULL;
        }        
    }
};
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  • Populating Next Right Pointers in Each Node II
    • Desicription
      • Solution
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