[数据结构与算法]算法打卡，用于自律

题目一

解法

class Solution {
    public TreeNode mirrorTree(TreeNode root) {
        if(root==null) return null;
        TreeNode node = new TreeNode(root.val);
        node.left = mirrorTree(root.right);
        node.right = mirrorTree(root.left);
        return node;
    }
}

题目二

?解法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    List<List<Integer>> res = new ArrayList<List<Integer>>();
    public List<List<Integer>> levelOrder(TreeNode root) {
        method(0,root);
        return res;
    }
    public void method(int depth, TreeNode root){
        if(root==null) return;
        if (res.size() == depth) {
            res.add(new ArrayList<Integer>());
        }
        res.get(depth).add(root.val);
        method(depth+1,root.left);
        method(depth+1,root.right);
    }
}

题目三

?解法一

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    ArrayList<Double> list = new ArrayList<Double>();
    ArrayList<List<Integer>> temp = new ArrayList<List<Integer>>();
    public List<Double> averageOfLevels(TreeNode root) {
        method(0,root);
        Double cont = 0.0;
        for(int i = 0;i<temp.size();i++){
            for(int j =0;j<temp.get(i).size();j++){
                cont+=temp.get(i).get(j);
            }
            Double d = cont/temp.get(i).size();
            list.add(d);
            cont = 0.0;
        }
        return list;
    }
    public void method(int depth,TreeNode root){
        if(root==null) return;
        if(temp.size()==depth){
            temp.add(new ArrayList<Integer>());
        }
        temp.get(depth).add(root.val);
        method(depth+1,root.left);
        method(depth+1,root.right);
    }
}

题目四

?解法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    List<Integer> list = new ArrayList<Integer>();
    public int kthLargest(TreeNode root, int k) {
        method(root);
        return list.get(list.size()-k);
    }
    public void method(TreeNode root){
        if(root==null) return;
        method(root.left);
        list.add(root.val);
        method(root.right);
    }
}