二叉树相关oj题

gywl2024-04-13 23:30

目录

检查两棵树是否相同

另一颗树的子树

翻转二叉树

判断一颗二叉树是否是平衡二叉树

对称二叉树

二叉树的构建及遍历

二叉树的分层遍历

[给定一个二叉树, 找到该树中两个指定节点的最近公共祖先](#给定一个二叉树, 找到该树中两个指定节点的最近公共祖先)

根据一棵树的前序遍历与中序遍历构造二叉树

根据一棵树的中序遍历与后序遍历构造二叉树

二叉树创建字符串

检查两棵树是否相同

java 复制代码 
/**
 * 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 {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        // 1.一个为空 一个不为空[结构上不一样]
        if (p != null && q == null || p == null && q != null) {
            return false;
        }

        // 2.第二步走完之后，要么都为空，要么都不为空 两个都是空
        if (p == null && q == null) {
            return true;
        }

        // 3.都不为空
        if (p.val != q.val) {
            return false;
        }

        // 4.此时代表2个都不为空 同时val值也是一样的
        // 5.说明根节点相同，接下来判断两棵树的左和两棵树的右是不是同时相同
        return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    }
}

另一颗树的子树

java 复制代码 
/**
 * 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 {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        // 1.一个为空 一个不为空[结构上不一样]
        if (p != null && q == null || p == null && q != null) {
            return false;
        }

        // 2.第二步走完之后，要么都为空，要么都不为空 两个都是空
        if (p == null && q == null) {
            return true;
        }

        // 3.都不为空
        if (p.val != q.val) {
            return false;
        }

        // 4.此时代表2个都不为空 同时val值也是一样的
        // 5.说明根节点相同，接下来判断两棵树的左和两棵树的右是不是同时相同
        return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    }

    public boolean isSubtree(TreeNode root, TreeNode subRoot) {
        if (root == null) {
            return false;
        }
        if (isSameTree(root, subRoot)) {
            return true;
        }
        if (isSubtree(root.left, subRoot)) {
            return true;
        }
        if (isSubtree(root.right, subRoot)) {
            return true;
        }
        return false;
    }
}

翻转二叉树

java 复制代码 
/**
 * 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 {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) {
            return null;
        }
        //减少一些递归和交换的次数
        if (root.left == null && root.right == null) {
            return root;
        }
        TreeNode tmp = root.left;
        root.left = root.right;
        root.right = tmp;
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }
}

判断一颗二叉树是否是平衡二叉树

java 复制代码 
/**
 * 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 {
    public boolean isBalanced(TreeNode root) {
        if (root == null)
            return true;

        int leftH = maxDepth(root.left);
        int rightH = maxDepth(root.right);

        return Math.abs(leftH - rightH) <= 1 && isBalanced(root.left) && isBalanced(root.right);
    }

    public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftHeight = maxDepth(root.left);
        int rightHeight = maxDepth(root.right);

        return leftHeight > rightHeight ? leftHeight + 1 : rightHeight + 1;
    }
}

但这样会有重复的计算，时间复杂度高，可优化为：

java 复制代码 
/**
 * 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 {
    public boolean isBalanced(TreeNode root) {
        if (root == null)
            return true;

        return maxDepth(root) >= 1;
    }

    public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftHeight = maxDepth(root.left);
        if (leftHeight < 0) {
            return -1;
        }
        int rightHeight = maxDepth(root.right);
        if (rightHeight < 0) {
            return -1;
        }
        if (Math.abs(leftHeight - rightHeight) <= 1) {
            return Math.max(leftHeight, rightHeight) + 1;
        } else {
            return -1;
        }
    }
}

对称二叉树

java 复制代码 
/**
 * 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 {
    public boolean isSymmetric(TreeNode root) {
        if (root == null)
            return true;
        return isSymmetricChild(root.left, root.right);
    }

    public boolean isSymmetricChild(TreeNode leftTree, TreeNode rightTree) {
        // 1.检查结构是否相同->一个为空 一个不为空
        if (leftTree != null && rightTree == null || leftTree == null && rightTree != null) {
            return false;
        }
        // 2.检查结构是否相同->处理【两个都为空】
        if (leftTree == null && rightTree == null) {
            return true;
        }
        // 3.检查结构是否相同->处理【两个都不为空,判断值一样吗】
        if (leftTree.val != rightTree.val) {
            return false;
        }
        // 4.此时两个引用都不为空 同时节点的值一样
        // 5.开始判断是否对称
        // 6.满足左子树的左和右子树的右对称 同时左子树的右和右子树的左对称
        return isSymmetricChild(leftTree.left, rightTree.right) && isSymmetricChild(leftTree.right, rightTree.left);
    }
}

二叉树的构建及遍历

java 复制代码 
import java.util.Scanner;
class TreeNode {
    public char val;
    public TreeNode left;
    public TreeNode right;

    public TreeNode(char val) {
        this.val = val;
    }
}


// 注意类名必须为 Main, 不要有任何 package xxx 信息
public class Main {

    public static int i = 0;
    public static TreeNode createTree(String str) {
        TreeNode root = null;
        if (str.charAt(i) != '#') {
            root = new TreeNode(str.charAt(i));
            i++;
            root.left = createTree(str);
            root.right = createTree(str);
        } else {
            i++;
        }
        return root;
    }

    public static void inorder(TreeNode root) {
        if (root == null) {
            return;
        }
        inorder(root.left);
        System.out.print(root.val + " ");
        inorder(root.right);
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        // 注意 hasNext 和 hasNextLine 的区别
        while (in.hasNextLine()) { // 注意 while 处理多个 case
            String str = in.nextLine();
            TreeNode root = createTree(str);
            inorder(root);
        }
    }
}

二叉树的分层遍历

java 复制代码 
/**
 * 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 {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> ret = new ArrayList<>();
        if (root == null) {
            return ret;
        }
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> list = new ArrayList<>();// 每层
            while (size > 0) {
                TreeNode cur = queue.poll();
                list.add(cur.val);
                if (cur.left != null) {
                    queue.offer(cur.left);
                }
                if (cur.right != null) {
                    queue.offer(cur.right);
                }
                size--;
            }
            ret.add(list);
        }
        return ret;
    }
}

给定一个二叉树, 找到该树中两个指定节点的最近公共祖先

java 复制代码 
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 * int val;
 * TreeNode left;
 * TreeNode right;
 * TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return null;
        }
        if (root == p || root == q) {
            return root;
        }
        TreeNode leftTree = lowestCommonAncestor(root.left, p, q);
        TreeNode rightTree = lowestCommonAncestor(root.right, p, q);
        // 两边都不为空
        if (leftTree != null && rightTree != null) {
            return root;
        } else if (leftTree != null) {
            // 左边不为空
            return leftTree;
        } else {
            // 右边不为空
            return rightTree;
        }
    }
}
java 复制代码 
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 * int val;
 * TreeNode left;
 * TreeNode right;
 * TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null) {
            return root;
        }
        // 1.获取root到指定节点路径上的所有结点
        Stack<TreeNode> stackP = new Stack<>();
        getPath(root, p, stackP);

        Stack<TreeNode> stackQ = new Stack<>();
        getPath(root, q, stackQ);

        // 2.上述代码执行完成 对应栈当中 存储了指定路径上的所有节点

        int sizeP = stackP.size();
        int sizeQ = stackQ.size();

        if (sizeP > sizeQ) {
            int size = sizeP - sizeQ;
            while (size != 0) {
                stackP.pop();
                size--;
            }
        } else {
            int size = sizeQ - sizeP;
            while (size != 0) {
                stackQ.pop();
                size--;
            }
        }
        // 3.上述if语句执行完成之后 两个栈当中的元素个数是一样的
        while (!stackP.isEmpty() && !stackQ.isEmpty()) {
            if (stackP.peek() == stackQ.peek()) {
                return stackP.peek();
            } else {
                stackP.pop();
                stackQ.pop();
            }
        }
        return null;
    }

    public boolean getPath(TreeNode root, TreeNode node, Stack<TreeNode> stack) {
        if (root == null) {
            return false;
        }
        stack.push(root);
        if (root == node) {
            return true;
        }
        boolean flg = getPath(root.left, node, stack);
        if (flg) {
            return true;
        }
        boolean flg2 = getPath(root.right, node, stack);
        if (flg2) {
            return true;
        }
        stack.pop();
        return false;
    }
}

根据一棵树的前序遍历与中序遍历构造二叉树

java 复制代码 
/**
 * 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 {

    public int preIndex;

    public TreeNode buildTree(int[] preorder, int[] inorder) {

        return buildTreeChilde(preorder, inorder, 0, inorder.length - 1);
    }

    private TreeNode buildTreeChilde(int[] preorder, int[] inorder, int inBegin, int inEnd) {
        if (inBegin > inEnd) {
            return null;
        }
        TreeNode root = new TreeNode(preorder[preIndex]);

        int rootIndex = findRootIndex(inorder, inBegin, inEnd, preorder[preIndex]);

        preIndex++;

        root.left = buildTreeChilde(preorder, inorder, inBegin, rootIndex - 1);
        root.right = buildTreeChilde(preorder, inorder, rootIndex + 1, inEnd);

        return root;
    }

    private int findRootIndex(int[] inorder, int inBegin, int inEnd, int key) {
        for (int i = inBegin; i <= inEnd; i++) {
            if (inorder[i] == key) {
                return i;
            }
        }
        return -1;
    }
}

根据一棵树的中序遍历与后序遍历构造二叉树

java 复制代码 
/**
 * 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 {
    public int postIndex = 0;

    public TreeNode buildTree(int[] inorder, int[] postorder) {
        postIndex = postorder.length - 1;
        return buildTreeChild(inorder, postorder, 0, inorder.length - 1);
    }

    private TreeNode buildTreeChild(int[] inorder, int[] postorder, int inBegin, int inEnd) {
        if (inBegin > inEnd) {
            return null;
        }

        TreeNode root = new TreeNode(postorder[postIndex]);

        int rootIndex = findRootIndex(inorder, inBegin, inEnd, postorder[postIndex]);

        postIndex--;

        root.right = buildTreeChild(inorder, postorder, rootIndex + 1, inEnd);
        root.left = buildTreeChild(inorder, postorder, inBegin, rootIndex - 1);

        return root;
    }

    private int findRootIndex(int[] inorder, int inBegin, int inEnd, int key) {
        for (int i = inBegin; i <= inEnd; i++) {
            if (inorder[i] == key) {
                return i;
            }
        }
        return -1;
    }
}

二叉树创建字符串

java 复制代码 
/**
 * 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 {
    public String tree2str(TreeNode root) {
        StringBuilder stringBuilder = new StringBuilder();

        tree2strChild(root, stringBuilder);

        return stringBuilder.toString();
    }

    private void tree2strChild(TreeNode root, StringBuilder stringBuilder) {
        if (root == null)
            return;
        stringBuilder.append(root.val);
        if (root.left != null) {
            stringBuilder.append("(");
            tree2strChild(root.left, stringBuilder);
            stringBuilder.append(")");
        } else {
            // 左边为空 右边也为空
            if (root.right == null) {
                return;
            } else {
                stringBuilder.append("()");
            }
        }
        // 处理root右树的情况
        if (root.right == null) {
            return;
        } else {
            stringBuilder.append("(");
            tree2strChild(root.right, stringBuilder);
            stringBuilder.append(")");
        }
    }
}
相关推荐
2的n次方_2 分钟前
二维费用背包问题
java·算法·动态规划
皮皮林5512 分钟前
警惕！List.of() vs Arrays.asList()：这些隐藏差异可能让你的代码崩溃！
java
莳光.2 分钟前
122、java的LambdaQueryWapper的条件拼接实现数据sql中and (column1 =1 or column1 is null)
java·mybatis
程序猿麦小七7 分钟前
基于springboot的景区网页设计与实现
java·spring boot·后端·旅游·景区
weisian15114 分钟前
认证鉴权框架SpringSecurity-2--重点组件和过滤器链篇
java·安全
蓝田～15 分钟前
SpringBoot-自定义注解，拦截器
java·spring boot·后端
.生产的驴18 分钟前
SpringCloud Gateway网关路由配置 接口统一 登录验证 权限校验 路由属性
java·spring boot·后端·spring·spring cloud·gateway·rabbitmq
simple_ssn30 分钟前
【C语言刷力扣】1502.判断能否形成等差数列
c语言·算法·leetcode
v'sir31 分钟前
POI word转pdf乱码问题处理
java·spring boot·后端·pdf·word
寂静山林38 分钟前
UVa 11855 Buzzwords
算法
热门推荐
01玄机平台应急响应—webshell查杀02Ubuntu 20.04使用Livox mid 360 测试 FAST_LIO03【HarmonyOS】HUAWEI DevEco Studio 下载地址汇总04新版微信小程序获取用户手机号05文件或文件夹名称中有空格如何批量去除06组基轨迹建模 GBTM的介绍与实现（Stata 或 R）07【TC3xx芯片】TC3xx芯片电源管理系统PMS详解08优化手机性能，解决卡顿问题：关闭这3个微信开关，释放内存空间09基于YOLOv10深度学习的CT扫描图像肾结石智能检测系统【python源码+Pyqt5界面+数据集+训练代码】深度学习实战、目标检测10【Flutter】基于 GetX 实现 Dio 的生命周期自动化