LeetCode - Populating Next Right Pointers in Each Node I && II

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本文介绍了如何使用常量额外空间填充二叉树节点的next指针，包括完美二叉树和非完美二叉树两种情况。通过遍历上一层节点生成next指针，实现了节点之间的连接。

Populating Next Right Pointers in Each Node I

https://leetcode.com/problems/populating-next-right-pointers-in-each-node/

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

这道题就是利用上一层已经生成的next指针来遍历上一层的子节点，即当前层。另外，由于保证是完全二叉树，所以不需要考虑很多特殊情况：

public void connect(TreeLinkNode root){
        if(root == null) return;
        TreeLinkNode lasthead = root;
        TreeLinkNode pre = null;
        while(lasthead!=null){
            TreeLinkNode curhead = lasthead.left;
            TreeLinkNode lastcur = lasthead;
            while(lastcur!=null && curhead!=null){
                if(pre==null){
                    pre = lastcur.left;
                }
                else{
                    pre.next = lastcur.left;
                    pre = pre.next;
                }
                pre.next = lastcur.right;
                pre = pre.next;
                lastcur = lastcur.next;
            }
            lasthead = curhead;
            pre = null;
        }
    }

Populating Next Right Pointers in Each Node II

https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

这道题跟上一道题差不多，区别在于不保证是完全二叉树，因此要考虑子节点为null的情况。

public void connect(TreeLinkNode root){
        if(root==null) return;
        TreeLinkNode lasthead = root;
        TreeLinkNode curhead = null;
        TreeLinkNode pre = null;
        while(lasthead!=null){
            TreeLinkNode lastcur = lasthead;
            while(lastcur!=null){
                if(lastcur.left!=null){
                    if(curhead==null){
                        curhead=lastcur.left;
                        pre = lastcur.left;
                    }
                    else{
                        pre.next = lastcur.left;
                        pre = pre.next;
                    }
                }
                if(lastcur.right!=null){
                    if(curhead==null){
                        curhead=lastcur.right;
                        pre = lastcur.right;
                    }
                    else{
                        pre.next = lastcur.right;
                        pre = pre.next;
                    }
                }
                lastcur = lastcur.next;
            }
            lasthead = curhead;
            curhead = null;
            pre = null;
        }
    }