Populating Next Right Pointers in Each Node

二叉树节点连接实现

最新推荐文章于 2022-02-28 18:16:11 发布

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本文介绍了一种常数额外空间复杂度下连接完美二叉树节点的方法，包括两种实现方式，一种使用队列进行层次遍历，另一种通过双循环逐层设置节点的next指针。

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

时间复杂度都是O（N），因为肯定要把所有节点遍历一次。
方法1：
空间复杂度O(N), BFS遍历tree，用curpos和lastpos标识层次变化，在每层遍历中设置next节点；

方法2：
空间复杂度O(1),用两个while循环。第一个while循环所有层次，第二个while在每层里遍历所有节点，在合适的时机设置nextline和prevnode就ok了。

/**
 * 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;
        
        // 1
        /* 
        queue<TreeLinkNode*> tq;
        TreeLinkNode *prevnode = NULL;
        tq.push(root);
        int curpos = 0, lastpos = 1;
        while (!tq.empty()) {
            TreeLinkNode *cur = tq.front();
            tq.pop();
            curpos++;
            if (prevnode == NULL) prevnode = cur;
            else {
                prevnode->next = cur;
                prevnode = cur;
            }
            
            prevnode->next = NULL;
            if (cur->left != NULL) tq.push(cur->left);
            if (cur->right != NULL) tq.push(cur->right);
            if (curpos == lastpos) {
                lastpos += tq.size();
                prevnode = NULL;
            }
        }
        */
        
        // 2
        TreeLinkNode *cur = root;
        cur->next = NULL;
        while (cur != NULL) {
            TreeLinkNode *nextLine = NULL;
            TreeLinkNode *prevNode = NULL;
            while (cur != NULL) {
                if (cur->left != NULL) {
                    if (nextLine == NULL) nextLine = cur->left;
                    if (prevNode != NULL) {
                        prevNode->next = cur->left;
                    }
                    prevNode = cur->left;
                }
                
                if (cur->right != NULL) {
                    if (nextLine == NULL) nextLine = cur->right;
                    if (prevNode != NULL) {
                        prevNode->next = cur->right;
                    }
                    
                    prevNode = cur->right;
                }
                
                if (prevNode != NULL) prevNode->next = NULL;
                cur = cur->next;
            }
            
            cur = nextLine;
        }
    }
};