#leetcode#Populating Next Right Pointers in Each Node

来源：互联网发布：淘宝店铺规则编辑：程序博客网时间：2024/06/06 03:35

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

这题咋一看很复杂，其实也还是BFS的变体， level order traversal，只是在之前BFS的基础上加一个next指针的赋值操作，

难得一眼就看出思路来了，多做题果然就能照猫画虎啊。。。

这里又犯了一个错误，用了queue.size()去判断是否是当前行的最后一个node，明明都已经提前用 int size = queue.size()去保存这个长度了。。。

/** * Definition for binary tree with next pointer. * public class TreeLinkNode { *     int val; *     TreeLinkNode left, right, next; *     TreeLinkNode(int x) { val = x; } * } */public class Solution {    public void connect(TreeLinkNode root) {        if(root == null){            return;        }        LinkedList<TreeLinkNode> queue = new LinkedList<>();        queue.offer(root);        while(!queue.isEmpty()){            int size = queue.size();            for(int i = 0; i < size; i++){                TreeLinkNode curNode = queue.pop();                // if(queue.size() == 0){   queue.size()是在随时变化的啊。。。 用你循环之前设的size                if(i == size - 1){  //  i == size - 1标示是前面一行的最后一个                    curNode.next = null;                }else{                    curNode.next = queue.peek();                }                if(curNode.left != null){                    queue.offer(curNode.left);                }                if(curNode.right != null){                    queue.offer(curNode.right);                }            }        }    }}

上面用queue的解法空间复杂度是O（n），其实这个题每一行的next指针，已经把当前行的结点连成了一个queue，利用好这一点则只要用几个指针来回赋值就可以了，lastHead表示当前行第一个节点， curHead表示下一行的第一个节点，用pre.next来连接同一行的前后结点

空间复杂度为 O（1）

/** * Definition for binary tree with next pointer. * public class TreeLinkNode { *     int val; *     TreeLinkNode left, right, next; *     TreeLinkNode(int x) { val = x; } * } */public class Solution {    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 = curHead;                    }else{                        pre.next = lastCur.left;                        pre = pre.next;                    }                }                                if(lastCur.right != null){                    if(curHead == null){                        curHead = lastCur.right;                        pre = curHead;                    }else{                        pre.next = lastCur.right;                        pre = pre.next;                    }                }                                lastCur = lastCur.next;            }                        lastHead = curHead;            curHead = null;        }    }}

0 0