序列化二叉树&&二叉搜索树的第k个结点&&字符串变形

NC123 序列化二叉树

题目链接

1、解题思路

队列中不能push空值，所以一旦遇到空，可以push进去一个101，表示空值。

2、代码

import java.util.*;
/*
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {
    private int getDepth(TreeNode node) {
        if (node == null) {
            return 0;
        } else {
            int l = getDepth(node.left);
            int r = getDepth(node.right);
            return l >= r ? l + 1 : r + 1;
        }
    }

    String Serialize(TreeNode root) {
        if (root == null) {
            return "";
        }
        StringBuilder stringBuilder = new StringBuilder();
        Queue<TreeNode> queue = new ArrayDeque<>();
        queue.add(root);
        int depth = getDepth(root);
        for (int i = 1; i <= depth; i++) {
            int size = queue.size();
            for (int j = 0; j < size; j++) {
                TreeNode poll = queue.poll();
                if (poll.val == 101) {
                    stringBuilder.append("null,");
                    queue.add(new TreeNode(101));
                    queue.add(new TreeNode(101));
                } else {
                    stringBuilder.append(poll.val);
                    stringBuilder.append(",");
                    if (poll.left != null) {
                        queue.add(poll.left);
                    } else {
                        queue.add(new TreeNode(101));
                    }
                    if (poll.right != null) {
                        queue.add(poll.right);
                    } else {
                        queue.add(new TreeNode(101));
                    }
                }
            }
        }
        stringBuilder.deleteCharAt(stringBuilder.length() - 1);
        return stringBuilder.toString();
    }

    private TreeNode TreeNodeHelper(String[] strings, int index) {
        if (index >= strings.length || strings[index].equals("null")) {
            return null;
        } else {
            TreeNode node = new TreeNode(Integer.parseInt(strings[index]));
            node.left = TreeNodeHelper(strings, index * 2 + 1);
            node.right = TreeNodeHelper(strings, index * 2 + 2);
            return node;
        }
    }

    TreeNode Deserialize(String str) {
        if (str == null || str.length() == 0) {
            return null;
        }
        String[] split = str.split(",");
        return TreeNodeHelper(split, 0);
    }
}

NC81 二叉搜索树的第k个结点

题目链接

1、解题思路

采用先序，然后每遍历一个结点，k值减1，输出k减到0的那个节点就可了。

2、代码

/*
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;

    public TreeNode(int val) {
        this.val = val;

    }

}
*/
public class Solution {
    private int k;

    private TreeNode preOrder(TreeNode pRoot) {
        if (pRoot == null) {
            return null;
        }
        TreeNode node = preOrder(pRoot.left);
        if (node != null) {
            return node;
        }
        this.k--;
        if (this.k == 0) {
            return pRoot;
        }
        TreeNode node1 = preOrder(pRoot.right);
        if (node1 != null) {
            return node1;
        } else {
            return null;
        }
    }

    TreeNode KthNode(TreeNode pRoot, int k) {
        this.k = k;
        return preOrder(pRoot);
    }
}

NC89 字符串变形

题目链接

1、解题思路

普通字符串处理。

2、代码

import java.util.*;

public class Solution {
    private String convert(String s) {
        StringBuilder stringBuilder = new StringBuilder(s).reverse();
        char[] chars = stringBuilder.toString().toCharArray();
        for (int i = 0; i < chars.length; i++) {
            if (chars[i] >= 'a' && chars[i] <= 'z') {
                chars[i] -= 32;
            } else {
                chars[i] += 32;
            }
        }
        return new String(chars);
    }

    public String trans(String s, int n) {
        StringBuilder stringBuilder = new StringBuilder();
        s = new StringBuilder(s).reverse().toString();
        int index = 0;
        int len = s.length();
        while (index < len) {
            if (s.charAt(index) == ' ') {
                stringBuilder.append(' ');
                index++;
            } else {
                int start = index;
                while (index < len && s.charAt(index) != ' ') {
                    index++;
                }
                stringBuilder.append(convert(s.substring(start, index)));
            }
        }
        return stringBuilder.toString();
    }
}