LeetCode 116:Populating Next Right Pointers in Each Node

Datetime:2016-08-23 01:23:07          Topic: LeetCode           Share

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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//对二叉树进行先序遍历,根据题述,给定的是满二叉树,左孩子为空,则右孩子一定为空,所以左孩子为空,则return
//如果左孩子不为空,则右孩子一定不为空,所以链接左孩子和右孩子即可(左孩子的next赋值为右孩子);
//由于先序遍历,所以父节点的next比子节点的next先被设置,故父节点不同的两个子节点进行连接,就可以用到父节点的next,
//即root->right->next = root->next->left;
class Solution {
public:
	void connect(TreeLinkNode *root) {
		if (root == NULL || root->left==NULL)
			return;

		if (root->left != NULL)
			root->left->next = root->right;

		if (root->right != NULL && root->next != NULL)
			root->right->next = root->next->left;
		else if (root->right != NULL)
			root->right->next = NULL;

		connect(root->left);
		connect(root->right);
	}
};




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