Leetcode 树 Populating Next Right Pointers in Each Node

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Populating Next Right Pointers in Each Node

Total Accepted: 13657 Total
Submissions: 39540

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

题意：给定一棵perfect binary tree，将它每一个节点的next指针都指向该节点右边的节点

思路：dfs

在connect一棵树的时候，需要知道这棵树的根节点和它右边的节点

1.将树的根节点和它右边的节点连接起来

2.递归地将左子树connect起来，需要知道左子树节点和右子树节点

3.递归地将右子树connect起来，需要知道右子树节点和根右边的节点的左子树节点

递归函数为：

void connect(TreeLinkNode *root, TreeLinkNode *sibling)

表示将以root为根的树和它的兄弟树sibling连接起来

复杂度：时间O(n)， 空间O(log n)

相关题目：

Populating Next Right Pointers in Each Node II

/**
 * 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){
    	connect(root, NULL);
    }

    void connect(TreeLinkNode *root, TreeLinkNode *sibling){
    	if(!root) return ;
    	root->next = sibling;
    	connect(root->left, root->right);
    	connect(root->right, sibling ? sibling->left : NULL);
    }
};

Leetcode 树 Populating Next Right Pointers in Each Node