【java-数据结构】别再死磕理论！这些 Java 二叉树题目带你快速上手

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前言

在Java编程的世界里，许多初学者常常陷入理论的迷宫，对着复杂的二叉树理论死磕，却忽略了实践的力量。其实，理论固然重要，但通过实际的题目练习，能更高效地掌握二叉树的相关知识。别再让晦涩的理论束缚你的脚步，那些看似复杂的二叉树概念，在实际题目中往往能变得清晰易懂。这里精心挑选了一系列Java二叉树题目，从基础的遍历到复杂的算法应用，让你在实战中快速上手，轻松跨越理论与实践的鸿沟，开启二叉树编程的新旅程。

1、检查两棵树是否相同。—题目链接

题目详解代码

package Demo2_25;/** * Created with IntelliJ IDEA. * Description: * User：Lenovo * Date:2025-02-25 * Time:19:13 */// 定义二叉树节点类class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; }}class Solution { public boolean isSameTree(TreeNode p, TreeNode q) { if ((p == null && q != null) || (p != null && q == null)) { return false; } if (p == null && q == null) { return true; } if (p.val != q.val) { return false; } return isSameTree(p.left, q.left) && isSameTree(p.right, q.right); } public static void main(String[] args) { // 创建第一棵二叉树 TreeNode p = new TreeNode(1); p.left = new TreeNode(2); p.right = new TreeNode(3); // 创建第二棵相同的二叉树 TreeNode q = new TreeNode(1); q.left = new TreeNode(2); q.right = new TreeNode(3); // 创建 Solution 类的实例 Solution solution = new Solution(); // 调用 isSameTree 方法判断两棵树是否相同 boolean result = solution.isSameTree(p, q); // 输出结果 System.out.println(\"两棵树是否相同: \" + result); // 创建一棵不同的二叉树 TreeNode r = new TreeNode(1); r.left = new TreeNode(3); r.right = new TreeNode(2); // 再次调用 isSameTree 方法判断 boolean differentResult = solution.isSameTree(p, r); System.out.println(\"两棵树是否相同: \" + differentResult); }}

2、另⼀棵树的⼦树。—题目链接

题目详解代码：

package Demo2_25;/** * Created with IntelliJ IDEA. * Description: * User：Lenovo * Date:2025-02-25 * Time:19:49 */// 定义二叉树节点类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 Solution1 { 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; } public boolean isSameTree(TreeNode p, TreeNode q) { if ((p == null && q != null) || (p != null && q == null)) { return false; } if (p == null && q == null) { return true; } if (p.val != q.val) { return false; } return isSameTree(p.left, q.left) && isSameTree(p.right, q.right); } public static void main(String[] args) { // 创建主树 TreeNode root = new TreeNode(3); root.left = new TreeNode(4); root.right = new TreeNode(5); root.left.left = new TreeNode(1); root.left.right = new TreeNode(2); // 创建子树 TreeNode subRoot = new TreeNode(4); subRoot.left = new TreeNode(1); subRoot.right = new TreeNode(2); // 创建 Solution 类的实例 Solution1 solution = new Solution1(); // 调用 isSubtree 方法判断主树中是否包含子树 boolean result = solution.isSubtree(root, subRoot); // 输出结果 System.out.println(\"主树中是否包含子树: \" + result); }}

3、翻转⼆叉树。 —题目链接

题目详解代码

package Demo2_25;/** * Created with IntelliJ IDEA. * Description: * User：Lenovo * Date:2025-02-25 * Time:19:58 */// 定义二叉树节点类class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; }}class Solution2 { public TreeNode invertTree(TreeNode root) { if (root == null) { return null; } 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; } public static void main(String[] args) { // 创建一棵二叉树 TreeNode root = new TreeNode(4); root.left = new TreeNode(2); root.right = new TreeNode(7); root.left.left = new TreeNode(1); root.left.right = new TreeNode(3); root.right.left = new TreeNode(6); root.right.right = new TreeNode(9); // 创建 Solution 类的实例 Solution2 solution = new Solution2(); // 调用 invertTree 方法翻转二叉树 TreeNode invertedRoot = solution.invertTree(root); // 简单输出翻转后二叉树的根节点值 if (invertedRoot != null) { System.out.println(\"翻转后二叉树的根节点值为: \" + invertedRoot.val); } else { System.out.println(\"翻转后的二叉树为空。\"); } }}

4、判断⼀棵⼆叉树是否是平衡⼆叉树。—题目链接

题目详解代码

package Demo2_25;/** * Created with IntelliJ IDEA. * Description: * User：Lenovo * Date:2025-02-25 * Time:20:28 */// 定义二叉树节点类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 Solution3 { public boolean isBalanced(TreeNode root) { //时间复杂度为N^2// if (root == null) {// return true;// }// int leftHeight = maxDepth(root.left);// int rightHeight = maxDepth(root.right);//// return Math.abs(leftHeight - rightHeight) <= 1// && isBalanced(root.left)// && isBalanced(root.right);// }//// public int maxDepth(TreeNode root) {// if (root == null) {// return 0;// }// int leftH = maxDepth(root.left);// int rightH = maxDepth(root.right);// return leftH > rightH ? leftH + 1 : rightH + 1;// } //时间复杂度为N if(root==null){ return true; } return maxDepth(root)>=0; } public int maxDepth(TreeNode root){ if(root==null){ return 0; } int leftH=maxDepth(root.left); if(leftH==-1){ return -1; } int rightH=maxDepth(root.right); if(leftH>=0&&rightH>=0&&Math.abs(leftH-rightH)<=1){ return Math.max(leftH,rightH)+1; } else{ return -1; } } public static void main(String[] args) { // 创建一个平衡二叉树 TreeNode balancedRoot = new TreeNode(3); balancedRoot.left = new TreeNode(9); balancedRoot.right = new TreeNode(20); balancedRoot.right.left = new TreeNode(15); balancedRoot.right.right = new TreeNode(7); // 创建一个不平衡二叉树 TreeNode unbalancedRoot = new TreeNode(1); unbalancedRoot.left = new TreeNode(2); unbalancedRoot.right = new TreeNode(2); unbalancedRoot.left.left = new TreeNode(3); unbalancedRoot.left.right = new TreeNode(3); unbalancedRoot.left.left.left = new TreeNode(4); unbalancedRoot.left.left.right = new TreeNode(4); // 创建 Solution 类的实例 Solution3 solution = new Solution3(); // 测试平衡二叉树 boolean isBalanced1 = solution.isBalanced(balancedRoot); System.out.println(\"第一棵树是否为平衡二叉树: \" + isBalanced1); // 测试不平衡二叉树 boolean isBalanced2 = solution.isBalanced(unbalancedRoot); System.out.println(\"第二棵树是否为平衡二叉树: \" + isBalanced2); }}

5、对称⼆叉树。—题目链接

题目详解代码

package Demo2_25;/** * Created with IntelliJ IDEA. * Description: * User：Lenovo * Date:2025-02-25 * Time:21:20 */// 定义二叉树节点类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 Solution4 { public boolean isSymmetric(TreeNode root) { if (root == null) { return true; } return isSymmetricChild(root.left, root.right); } public boolean isSymmetricChild(TreeNode leftTree, TreeNode rightTree) { if ((leftTree == null && rightTree != null) || (leftTree != null && rightTree == null)) { return false; } if (leftTree == null && rightTree == null) { return true; } if (leftTree.val != rightTree.val) { return false; } return isSymmetricChild(leftTree.left, rightTree.right) && isSymmetricChild(leftTree.right, rightTree.left); } public static void main(String[] args) { // 创建一个对称的二叉树 TreeNode symmetricRoot = new TreeNode(1); symmetricRoot.left = new TreeNode(2); symmetricRoot.right = new TreeNode(2); symmetricRoot.left.left = new TreeNode(3); symmetricRoot.left.right = new TreeNode(4); symmetricRoot.right.left = new TreeNode(4); symmetricRoot.right.right = new TreeNode(3); // 创建一个不对称的二叉树 TreeNode asymmetricRoot = new TreeNode(1); asymmetricRoot.left = new TreeNode(2); asymmetricRoot.right = new TreeNode(2); asymmetricRoot.left.right = new TreeNode(3); asymmetricRoot.right.right = new TreeNode(3); Solution4 solution = new Solution4(); // 测试对称二叉树 boolean isSymmetric1 = solution.isSymmetric(symmetricRoot); System.out.println(\"对称二叉树测试结果: \" + isSymmetric1); // 测试不对称二叉树 boolean isSymmetric2 = solution.isSymmetric(asymmetricRoot); System.out.println(\"不对称二叉树测试结果: \" + isSymmetric2); }}

6、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 。—题目链接

题目详解代码

package Demo2_26;import java.util.Stack;/** * Created with IntelliJ IDEA. * Description: * User：Lenovo * Date:2025-02-26 * Time:20:58 */// 定义二叉树节点类class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; }} //解法一：// 定义解决方案类//class Solution5 {// public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {// if (root == null) {// return null;// }// if (root == p || root == q) {// return root;// }// TreeNode leftT = lowestCommonAncestor(root.left, p, q);// TreeNode rightT = lowestCommonAncestor(root.right, p, q);// if (leftT != null && rightT != null) {// return root;// } else if (leftT != null) {// return leftT;// } else if (rightT != null) {// return rightT;// }// return null;// }  //解法二：  class Solution5 { 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; } flg=getPath(root.right,node,stack); if(flg){ return true; } stack.pop(); return false; } public TreeNode lowestCommonAncestor(TreeNode root,TreeNode p,TreeNode q){ if(root==null){ return null; } Stack<TreeNode> stack1=new Stack<>(); Stack<TreeNode> stack2=new Stack<>(); getPath(root,p,stack1); getPath(root,q,stack2); int size1=stack1.size(); int size2=stack2.size(); int size=size1-size2; if(size<0){ size=Math.abs(size); while(size!=0){ stack2.pop(); size--; } } else{ while(size!=0){ stack1.pop(); size--; } } while(!stack1.isEmpty()&&!stack2.isEmpty()){ if(stack1.peek()==stack2.peek()){ return stack1.pop(); }else{ stack1.pop(); stack2.pop(); } } return null; } // 主类，用于测试 public static void main(String[] args) { // 构建一个简单的二叉树 TreeNode root = new TreeNode(3); root.left = new TreeNode(5); root.right = new TreeNode(1); root.left.left = new TreeNode(6); root.left.right = new TreeNode(2); root.right.left = new TreeNode(0); root.right.right = new TreeNode(8); root.left.right.left = new TreeNode(7); root.left.right.right = new TreeNode(4); // 定义要查找的两个节点 TreeNode p = root.left; // 节点 5 TreeNode q = root.right; // 节点 1 // 创建解决方案对象 Solution5 solution = new Solution5(); // 调用 lowestCommonAncestor 方法查找最近公共祖先 TreeNode lca = solution.lowestCommonAncestor(root, p, q); // 输出结果 if (lca != null) { System.out.println(\"最近公共祖先是: \" + lca.val); } else { System.out.println(\"未找到最近公共祖先。\"); } }}

7、⼆叉树创建字符串。—题目链接

题目详解代码

package Demo2_27;/** * Created with IntelliJ IDEA. * Description: * User：Lenovo * Date:2025-02-27 * Time:17:14 */// 定义二叉树节点类class TreeNode { int val; TreeNode left; TreeNode right; // 构造函数，用于初始化节点的值 TreeNode(int x) { val = x; }}class Solution { // 主方法，将二叉树转换为字符串 public String tree2str(TreeNode root) { StringBuilder stringBuilder = new StringBuilder(); tree2strChild(root, stringBuilder); return stringBuilder.toString(); } // 辅助方法，递归地将二叉树的节点添加到字符串构建器中 public void tree2strChild(TreeNode root, StringBuilder stringBuilder) { if (root == null) { return; } // 添加当前节点的值 stringBuilder.append(root.val); if (root.left != null) { // 若左子节点不为空，添加左括号 stringBuilder.append(\"(\"); // 递归处理左子树 tree2strChild(root.left, stringBuilder); // 添加右括号 stringBuilder.append(\")\"); } else { if (root.right == null) { return; } else { // 左子节点为空但右子节点不为空，添加空括号 stringBuilder.append(\"()\"); } } if (root.right != null) { // 若右子节点不为空，添加左括号 stringBuilder.append(\"(\"); // 递归处理右子树 tree2strChild(root.right, stringBuilder); // 添加右括号 stringBuilder.append(\")\"); } else { return; } } public static void main(String[] args) { // 创建示例二叉树 TreeNode root = new TreeNode(1); root.left = new TreeNode(2); root.right = new TreeNode(3); root.left.left = new TreeNode(4); // 创建 Solution 类的实例 Solution solution = new Solution(); // 调用 tree2str 方法将二叉树转换为字符串 String result = solution.tree2str(root); // 输出结果 System.out.println(result); }}

Java二叉树的题目就分享到这里了