Morris Traversal方法遍历二叉树(非递归,不用栈,O(1)空间) /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ import java.uti
}publicstaticvoidmain(String[] args){BinaryTreeInOrderTraversalbinaryTreeInOrderTraversal=newBinaryTreeInOrderTraversal();char[] arr1 =newchar[]{'1','#','2','3'};char[] arr2 =newchar[]{'1','2','3','#','#','4','#','#','5'}; System.out.println(Arrays.toString(binaryTreeI...
2.2 last.right 不为 null,说明之前已经访问过,第二次来到这里,表明当前子树遍历完成,保存 cur 的值,更新 cur = cur.right public List<Integer> inorderTraversal3(TreeNode root) { List<Integer> ans = new ArrayList<>(); TreeNode cur = root; while (cur != null) { //情况 1 if (cur.left ...
private void pushAllTheLeft(Stack<TreeNode> s, TreeNode root){ s.push(root); while(root.left!=null){ root = root.left; s.push(root); } } } Binary Tree Postorder Traversal Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,...
Leetcode 94: Binary Tree Inorder Traversal-中序遍历 自动驾驶搬砖师 一步一个台阶,一天一个脚印 一、题目描述 给定一个二叉树,返回它的中序遍历。 二、解题思路 2.1 递归法 时间复杂度 O(n) 空间复杂度 O(n) 2.2 迭代法 时间复杂度 O(n) 空间复杂度 O(n)(1) 同理创建一个Stack,然后按左/中/右...
Given preorder and inorder traversal of a tree, construct the binary tree. 二分法 复杂度 时间O(N^2) 空间 O(N) 思路 我们先考察先序遍历序列和中序遍历序列的特点。对于先序遍历序列,根在最前面,后面部分存在一个分割点,前半部分是根的左子树,后半部分是根的右子树。对于中序遍历序列,根在中间部分...
The inOrder traversal is one of the most popular ways to traverse a binary tree data structure in Java. TheinOrdertraversal is one of the three most popular ways to traverse a binary tree data structure, the other two being thepreOrderandpostOrder. During the in-order traversal algorithm, ...
二叉树中序遍历的非递归算法和前序遍历思想差不多都是使用栈结构进行求解,参考Leetcode: Binary Tree Preorder Traversal(二叉树前序遍历)。 非递归解法(C++): class Solution { public: vector<int> inorderTraversal(TreeNode *root) { vector<int> result; ...
Can you solve this real interview question? Binary Tree Preorder Traversal - Given the root of a binary tree, return the preorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [1,2,3] Explanation: [https://assets.l
* TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{public List<Integer>inorderTraversal(TreeNode root){ArrayList<Integer>result=newArrayList<Integer>();if(root==null)returnresult;inorderTraversal(root,result);returnresult;}privatevoidinorderTraversal(TreeNode root,Arra...