题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 根据前序遍历和中序遍历结果构造二叉树。 思路分析: 分析二叉树前序遍历和中序遍历的结果我们发现: 二叉树前序遍历的第一个节点是根节点。 在中序遍历...
*/publicclassBinaryTreeInOrderTraversal{privateint[] result =null;intpos=0;publicint[] traversal(char[] tree) { result =newint[tree.length]; pos =0; traversalByRecursion(createTree(tree));returnresult; }publicint[] traversal1(char[] tree) { result =newint[tree.length]; pos =0; travers...
Val) // 接下来该处理 cur 的右子树 cur = cur.Right } return ans } 题目链接: Binary Tree Inorder Traversal: leetcode.com/problems/b 二叉树的中序遍历: leetcode.cn/problems/bi LeetCode 日更第 238 天,感谢阅读至此的你 欢迎点赞、收藏鼓励支持小满...
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...
inorder(root -> right,vec); } } vector<int> inorderTraversal(TreeNode* root) { vector<int> vec; if(root == NULL)return vec; inorder(root,vec); return vec; } }; 递归 class Solution { public: void inorder(TreeNode* root, vector<int> & vec) { if(root -> left != NULL) {...
Given a binary tree, return theinordertraversal of its nodes' values. 示例: 代码语言:javascript 代码运行次数:0 AI代码解释 输入:[1,null,2,3]1\2/3输出:[1,3,2] 进阶:递归算法很简单,你可以通过迭代算法完成吗? Follow up:Recursive solution is trivial, could you do it iteratively?
inorderTraversal(root.right); } return res; } } 时间复杂度:O(n),其中n为二叉树节点的个数。二叉树的遍历中每个节点会被访问一次且只会被访问一次。 空间复杂度:O(n)。空间复杂度取决于递归的栈深度,而栈深度在二叉树为一条链的情况下会达到O(n)的级别。
# class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: def inorderTraversal(self, root: TreeNode) -> List[int]: inorderList = [] stack = [] while stack or root: while root: stack.append(root) root = root.left...
ConstructBinaryTreefromPreorderandInorderTraversal Jan 30, 2015 ContainerWithMostWater.java Container With Most Water Dec 21, 2014 ConvertSortedArraytoBinarySearchTree.java add inorder solution Jan 29, 2015 ConvertSortedListtoBinarySearchTree.java ...
0094 Binary Tree Inorder Traversal Go 63.3% Medium 0095 Unique Binary Search Trees II Go 40.6% Medium 0096 Unique Binary Search Trees Go 52.9% Medium 0097 Interleaving String 31.5% Hard 0098 Validate Binary Search Tree Go 27.8% Medium 0099 Recover Binary Search Tree Go 39.6% Hard 010...