Populating Next Right Pointers in Each Node II (连接兄弟节点）

最新推荐文章于 2020-11-14 09:32:10 发布

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#C++ #algorithm #二叉树 #层次遍历

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本文介绍了一种解决二叉树层次遍历并连接同一层节点的问题的方法。使用vector模拟队列，实现节点间的next指针连接，适用于任意二叉树结构。

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,

After calling your function, the tree should look like:

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

给出一棵二叉树，要求把每一层的兄弟节点顺序连接起来。显然是层次遍历，利用vector模拟队列就可以了队列头从top开始，top == queue.size()表示队列元素已空。

/**
 * 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) {
		// IMPORTANT: Please reset any member data you declared, as
		// the same Solution instance will be reused for each test case.
		vector<TreeLinkNode*> queue;
		int top = -1;
		if (root == NULL) return;
		queue.push_back(root);
		++top;
		while (top != queue.size())
		{
			int len = queue.size();
			for (int i = top; i < len - 1; ++i)
			{
				queue[i]->next = queue[i + 1];
			}
			for (int i = top; i < len; ++i, ++top)
			{
				if (queue[i]->left) queue.push_back(queue[i]->left);
				if (queue[i]->right) queue.push_back(queue[i]->right);
			}
		}
	}
};