题目Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree [1,null,2,3], return [1,3,2]. Note: Recursive solution is trivial, could you do it iter…
}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...
inorderTraversal(root.right); } return list; } } 非递归实现: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<Integer> inorderTravers...
3.1 递归法 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<int> inorderTraversal(TreeNode* root) { vector<int...
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] ...
题目地址:https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/ 题目描述 Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. ...
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...
** Inorder Traversal: left -> root -> right ** Preoder Traversal: root -> left -> right ** Postoder Traveral: left -> right -> root 记忆方式:order的名字指的是root在什么位置。left,right的相对位置是固定的。 图片来源:https://leetcode.com/articles/binary-tree-right-side-view/ ...
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...
2019-12-21 22:15 −Description Flatten a binary tree to a fake "linked list" in pre-order traversal. Here we use the right pointer in TreeNode as the ... YuriFLAG 0 235 Mybatis:java.lang.IllegalArgumentException: invalid comparison: java.util.Date and java.lang.String ...