leetcode 117 Populating Next Right Pointers in Each Node II

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本文介绍了一种在二叉树中使用广度优先遍历填充每个节点的next指针的方法，使得每个节点的next指针指向其右侧的下一个节点。提供了两种解决方案，一种使用两个数组实现，另一种则在常数额外空间限制下进行优化。

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117 Populating Next Right Pointers in Each Node II 广度优先遍历

问题地址
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.
Recursive approach is fine, implicit stack space does not count as extra space for this problem.
Example:

Given the following binary tree
$1/ 2 3/ \ 4 5 7$
After calling your function, the tree should look like
函数结果图片

my first solution

采用两个数组实现广度优先遍历

public class Solution {
    public void connect(TreeLinkNode root) {
        List<TreeLinkNode> list1 = new LinkedList<TreeLinkNode>();
        List<TreeLinkNode> list2 = new LinkedList<TreeLinkNode>();
        List<TreeLinkNode> list;
        if(root == null) return;
        
        list1.add(root);
        int i = 0;
        while(list1.size()!=0)
        {
            i = 0;
            while(i < list1.size())
            {
                root = list1.get(i);
                if(i == list1.size()-1) root.next = null;
                else root.next = list1.get(i+1);
                if(root.left!=null)list2.add(root.left);
                if(root.right!=null)list2.add(root.right);
                i++;
            }
            list1.clear();
            list = list1;
            list1 = list2;
            list2 = list;
        }
    }
}

better solution

   //基于广度优先的遍历
    public void connect(TreeLinkNode root) {

        TreeLinkNode head = null; //下一层的第一个节点
        TreeLinkNode prev = null; //下一层当前遍历到的节点
        TreeLinkNode cur = root;  //当前遍历的节点

        while (cur != null) {
            while (cur != null) { //在当前层迭代，修改下一层节点的next指针，使其指向正确的节点
                //遍历当前节点的左孩子
                if (cur.left != null) {
                    if (prev != null) {
                         prev.next = cur.left;
                    } else {
                        head = cur.left;
                    }
                    prev = cur.left;
                }
                //遍历当前节点的右孩子
                if (cur.right != null) {
                    if (prev != null) {
                        prev.next = cur.right;
                    } else {
                        head = cur.right;
                    }
                    prev = cur.right;
                }
                //遍历当前层的下一个节点
                cur = cur.next;
            }
            
            //移到下一层节点开始遍历
            cur = head;
            head = null;
            prev = null;
        }
        
    }