闲来无事刷波力扣

首先我要吐槽下，一回家果然被父母各种拉出去了，又是见人又是逛街的，还时不时的提下我没有女朋友的事实，顺便吃下自己兄弟姐妹的狗粮，不过看到自己的堂哥们和我一样，我的内心就平静许多了，晚上简单用java写了两题力扣，发一下自己的代码吧。

105 从前序与中序遍历序列构造二叉树

根据一棵树的前序遍历与中序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

例如，给出

前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]

返回如下的二叉树：

    3
   / \
  9  20
    /  \
   15   7

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        
        if(preorder.length == 0){
            return null;
        }
        
        int head = preorder[0];
        TreeNode node = new TreeNode(head);
        int pos = 0;
        for(int i=0; i < inorder.length; i++){
            if(inorder[i] == head){
                pos = i;
            }
        }
        node.val = head;
        
        int leftprepart[] = Arrays.copyOfRange(preorder, 1, 1+pos);
        int leftinpart[] = Arrays.copyOfRange(inorder, 0, pos);
        int rightprepart[] = Arrays.copyOfRange(preorder, 1+pos, preorder.length);
        int rightinpart[] = Arrays.copyOfRange(inorder, 1+pos, inorder.length);
        
        node.left = buildTree(leftprepart, leftinpart);
        node.right = buildTree(rightprepart, rightinpart);
        
        return node;
    }
}

116 填充每个节点的下一个右侧节点指针

给定一个完美二叉树，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。

初始状态下，所有 next 指针都被设置为 NULL。

 

示例：

输入：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

输出：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}

解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。

 

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}
    
    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _left, Node _right, Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/
class Solution {
    List forenode = new ArrayList();
    
    public Node connect(Node root) {
        visitnode(root, 0);
        return root;
    }
    
    public void visitnode(Node root, int layer){
        if(root == null){
            return;
        }
        if(forenode.size() < layer+1){
            forenode.add(root);
        }else{
            forenode.get(layer).next = root;
            forenode.set(layer, root);
        }
        visitnode(root.left, layer + 1);
        visitnode(root.right, layer + 1);
    }
}