leetcode 116 Populating Next Right Pointers in Each Node

最新推荐文章于 2021-09-24 12:05:40 发布

walter1990

最新推荐文章于 2021-09-24 12:05:40 发布

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本文介绍了一种在完美二叉树中填充每个节点的 next 指针的方法，使其指向右侧相邻节点。若无相邻节点，则设置为 NULL。文章提供了一个解决方案，确保仅使用常数额外空间，并假设输入为完美二叉树。

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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,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

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/**
 * 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;

		while(true) {
			TreeLinkNode *node = root->left, *temp=root->left;
			if(node == NULL) break;
			int cnt=1;

			for(; root!=NULL; root=root->next) {
				if(cnt > 1) {
					node->next = root->left;
					node = node->next;
				}

				node->next = root->right;
				node = node->next;

				cnt++;
			}

			root=temp;
		}

		return;
    }
};