[LeetCode]:116：Populating Next Right Pointers in Each Node-优快云博客

本文介绍了一种在完美二叉树中填充每个节点的next指针的方法，使得每个节点指向其右侧相邻节点。提供了三种解决方案：深度优先搜索（DFS）、广度优先搜索（BFS）及常数额外空间的逐层遍历方法。

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题目：

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

思路1：使用深度优先搜索遍历

代码：

    public static void connect_DFS(TreeLinkNode root) {
        if(root != null){
            if(root.left != null && root.right != null){
                root.left.next = root.right;
                System.out.println(root.left.val +" Next : "+ root.right.val);
                if(root.left.right != null && root.right.left != null){
                    root.left.right.next =  root.right.left;
                    System.out.println(root.left.right.val +" Next : "+ root.right.left.val);
                }
            }
            
            if(root.left != null ){
                connect_DFS(root.left);
            }
            if(root.right != null ){
                connect_DFS(root.right);
            }
        }
    }

分析：存在缺点：第一左子树的最右边的节点的情况处理不到，分析貌似用深搜不行，后来看了网上高人的解答，发现：root.right.next = root.next.left

更改代码为：

public class Solution {
    public static void connect(TreeLinkNode root) {
        if(root != null){
            if(root.left != null && root.right != null){
                root.left.next = root.right;
                //System.out.println(root.left.val +" Next : "+ root.right.val);
                
                if(root.next!= null){
                    root.right.next =  root.next.left;
                    //System.out.println(root.right.val +" Next : "+ root.next.left.val);
                }
            }
            
            if(root.left != null ){
                connect(root.left);
            }
            if(root.right != null ){
                connect(root.right);
            }
        }
    }
}

思路二：采用广度优先搜索

代码：

/**
 * 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 static void connect(TreeLinkNode root) {
        if(root!= null){
            Queue<TreeLinkNode> queListFlag = new LinkedList<TreeLinkNode>();
            
            queListFlag.add(root);

            while(queListFlag.size()!=0){
                ArrayList arrayTemp=new ArrayList();

                while(queListFlag.size()!=0){
                    arrayTemp.add(queListFlag.peek());
                    queListFlag.remove();
                }
                
                if(arrayTemp.size()==1){
                    TreeLinkNode NodeTemp = (TreeLinkNode)arrayTemp.get(0);
                    
                    if(NodeTemp.left != null && NodeTemp.right != null){
                        NodeTemp.left.next = NodeTemp.right;
                        //System.out.println(NodeTemp.left.val +" Next : "+ NodeTemp.right.val);
                    }
                    
                    if(NodeTemp.left != null ){
                        queListFlag.add(root.left);
                    }
                    if(NodeTemp.right != null ){
                        queListFlag.add(root.right);
                    } 
                }else if(arrayTemp.size()>1){
                    for(int i=0;i<arrayTemp.size()-1;i++) {
                        TreeLinkNode NodeTemp1 = (TreeLinkNode)arrayTemp.get(i);
                        TreeLinkNode NodeTemp2 = (TreeLinkNode)arrayTemp.get(i+1);
                        
                        if(NodeTemp1.left != null && NodeTemp1.right != null){
                            NodeTemp1.left.next = NodeTemp1.right;
                            //System.out.println(NodeTemp1.left.val +" Next : "+ NodeTemp1.right.val);
                        }
                        
                        if(NodeTemp1.right != null && NodeTemp2.left != null){
                            NodeTemp1.right.next = NodeTemp2.left;
                            //System.out.println(NodeTemp1.right.val +" Next : "+ NodeTemp2.left.val);
                        }
                        
                        if(i == arrayTemp.size()-2){
                            if(NodeTemp2.left != null && NodeTemp2.right != null){
                                NodeTemp2.left.next = NodeTemp2.right;
                                //System.out.println(NodeTemp2.left.val +" Next : "+ NodeTemp2.right.val);
                            }
                        }

                    }
                    
                    for(int i=0;i<arrayTemp.size();i++) {
                        TreeLinkNode NodeTemp = (TreeLinkNode)arrayTemp.get(i);
                        if(NodeTemp.left != null ){
                            queListFlag.add(NodeTemp.left);
                        }
                        if(NodeTemp.right != null ){
                            queListFlag.add(NodeTemp.right);
                        } 
                    }
                }
            }
            
        }
    }
}

思路三：不采用其它的存储空间

题目要求：You may only use constant extra space.

所以逐层遍历节点的时候，需要将下一层节点的都连接上

抄袭高人的代码如下：

public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode levelHead = root, nextLevelHead = null;
        while (levelHead != null) {
            TreeLinkNode node = levelHead, tail = null;
            while (node != null) {
                if (node.left != null && node.right != null) {
                    node.left.next = node.right;
                }
                TreeLinkNode sub;
                if (node.left != null)
                    sub = node.left;
                else if (node.right != null)
                    sub = node.right;
                else
                    sub = null;
                if (sub != null) {
                    if (nextLevelHead == null) {
                        nextLevelHead = sub;
                        tail = sub;
                    } else {
                        tail.next = sub;
                    }
                    while (tail.next != null)
                        tail = tail.next;
                }
                node = node.next;
            }
            levelHead = nextLevelHead;
            nextLevelHead = null;
        }
    }
}