叫 Morris Traversal,在介绍这种方法之前,先来引入一种新型树,叫Threaded binary tree,这个还不太好翻译,第一眼看上去以为是叫线程二叉树,但是感觉好像又跟线程没啥关系,后来看到网上有人翻译为螺纹二叉树,但博主认为这翻译也不太敢直视,很容易让人联想到为计划生育做出突出贡献的某世界...
package leetcode; import java.util.*; //方法一 数据结构:树的中序遍历:1、递归 2、非递归 //始终操作的是根节点 ->( root=root.left),把右边节点移到根节点上进行; public class BinaryTreeInorderTraversal { private class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) {...
94. Binary Tree Inorder TraversalEasy Topics Companies Given the root of a binary tree, return the inorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [1,3,2] Explanation: Example 2: Input: root = [1,2,3,4,5,null,8,null,null,6,7,9] ...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tre...
原题链接:http://oj.leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/ 这道题是树中比較有难度的题目。须要依据先序遍历和中序遍历来构造出树来。这道题看似毫无头绪。事实上梳理一下还是有章可循的。以下我们就用一个样例来解释怎样构造出树。
The function returns the root of the constructed binary tree. 4. Time & Space Complexity Analysis: 4.1 Time Complexity: 4.1.1 Lists as parameters In each recursive call, the index() function is used to find the index of the root value in the inorder traversal list. This function has a ...
public List<Integer> inorderTraversal(TreeNode root) { List<Integer> ans = new ArrayList<>(); getAns(root, ans); return ans; } private void getAns(TreeNode node, List<Integer> ans) { if (node == null) { return; } getAns(node.left, ans); ans.add(node.val); getAns(node.right...
名字叫做, morrois traversal, 自己写了下: My code: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicList<Integer>inorderTraversal(TreeNoderoot){List...
Binary Tree Level Order Traversal 二叉树层序遍历 Example Givenbinary tree[3,9,20,null,null,15,7],3/\920/\157returnits level order traversalas:[[3],[9,20],[15,7]] BFS方法 var levelOrder=function(root){if(!root)return[]conststack=[root]constres=[]while(stack.length){constlen=stack...
代码如下: 1/**2* Definition for a binary tree node.3* public class TreeNode {4* int val;5* TreeNode left;6* TreeNode right;7* TreeNode(int x) { val = x; }8* }9*/10classSolution {11publicList<Integer>inorderTraversal(TreeNode root) {12List<Integer> res =newArrayList<Integer>...