BinaryTree bt = BinaryTree.create(); // traversing binary tree using InOrder traversal using recursion System.out .println("printing nodes of a binary tree on InOrder using recursion"); bt.inOrder(); System.out.println();// insert new line // traversing binary tree on InOrder traversal ...
classSolution(object):definorderTraversal(self, root):#递归""":type root: TreeNode :rtype: List[int]"""ifnotroot:return[]returnself.inorderTraversal(root.left) + [root.val] + self.inorderTraversal(root.right) Java解法 classSolution {publicList < Integer >inorderTraversal(TreeNode root) ...
1. 中序遍历(In-order Traversal):先访问左子树,然后是根节点,最后访问右子树。这种遍历方式可以按照“左-根-右”的顺序访问所有节点。 2. 前序遍历(Pre-order Traversal):先访问根节点,然后是左子树,最后访问右子树。这种遍历方式可以按照“根-左-右”的顺序访问所有节点。 3. 后序遍历(Post-order Traversal...
The easiest way to implement theinOrdertraversal algorithm in Java or any programming language is by using recursion. Since the binary tree is a recursive data structure, recursion is the natural choice for solving a tree-based problem. TheinOrder()method in theBinaryTreeclass implements the logi...
Morris方法可参考帖子:Morris Traversal方法遍历二叉树(非递归,不用栈,O(1)空间) Java: Without Recursion, using Stack 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 publicclassSolution { /** * @param root: The root of binary tree. ...
Reverse inorder traversal of the above tree is: 10 9 8 7 6 5 4 3 2 1 0 C++ implementation: #include <bits/stdc++.h>usingnamespacestd;classTreeNode{// tree node is definedpublic:intval; TreeNode*left; TreeNode*right; TreeNode(intdata) ...
add(root.val); inorderTraversal(root.right); } return list; } } 复杂度分析 时间复杂度: O(n) 空间复杂度: O(n) 2. 使用栈遍历 想要通过遍历实现必然需要使用到额外的空间,每次进入左子节点前记录当前的节点,当遍历到头的时候回退回来,因此这种 LIFO 的数据结构就非常适合使用栈: /** * ...
查询index部分可以使用Map做查询,建议和下一道题 leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal 中后序构造BST和leetcode 449. Serialize and Deserialize BST 二叉搜索树BST的序列化和反序列化一起学习。 建议和leetcode 87. Scramble String 字符串拼凑 && DFS深度优先搜索 和 leetco...
Similarly, we will search for 5 in the inorder traversal range.Figure 3: Step 3Further recursions:Figure 4: Step 4Figure 5: Step 5Figure 6: Step 6The final built tree is:Figure 7: Final constructed TreeC++ Implementation#include <bits/stdc++.h> using namespace std; // tre...
The recursion splits the problem into two halves at each level, and there are h levels in the binary tree. Therefore, the time complexity is O(n * h). 4.1.3 High Efficiency Indices as parameters The time complexity of constructing the binary tree using this method is O(n). The main ...