Preorder, Inorder, and Postorder Iteratively Summarization[1] 1.Pre Order Traverse public List<Integer> preorderTraversal(TreeNode root) { List<Integer> result = new ArrayList<>(); Deque<TreeNode> stack = new ArrayDeque<>();…

[144-Binary Tree Preorder Traversal(二叉树非递归前序遍历)] [LeetCode-面试算法经典-Java实现][全部题目文件夹索引] 原题 Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solut…

144. Binary Tree Preorder Traversal 前序的非递归遍历:用堆来实现 如果把这个代码改成先向堆存储左节点再存储右节点,就变成了每一行从右向左打印 如果用队列替代堆,并且先存储左节点,再存储右节点,就变成了逐行打印 class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> result; if(root == NULL) return res…

Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, given inorder = [9,3,15,20,7] postorder = [9,15,7,20,3] Return the following binary tree: 3 / \ 9 2…

Binary Tree Preorder Traversal 题目链接 题目要求: Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? 递…

LeetCode:Construct Binary Tree from Inorder and Postorder Traversal Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree.                                            …

LeetCode 原题链接 Construct Binary Tree from Inorder and Postorder Traversal - LeetCode Construct Binary Tree from Preorder and Postorder Traversal - LeetCode 题目大意 给定一棵二叉树的中序遍历和后序遍历,求这棵二叉树的结构. 给定一棵二叉树的前序遍历和中序遍历,求这棵二叉树的结构. 样例 Input: inorder = [9, 3, 15, 2…

Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 这道题之前算法课上好像遇到过,思路也很简单的. 思路:后序序列的最后一个元素就是树根,然后在中序序列中找到这个元素(由于题目保证没有相同的元素,因此可以唯一找到),中序序列中这个元素的左边就是左子树的中序,右边就是右子树的…

1.后序遍历的非递归实现.(左右根) 难点:后序遍历的非递归实现是三种遍历方式中最难的一种.因为在后序遍历中,要保证左孩子和右孩子都已被访问并且左孩子在右孩子前访问才能访问根结点,这就为流程的控制带来了难题.下面介绍两种思路. 思路:有个关键的就是unUsed这个标识符. 当unUsed=1时,表示该节点未遍历过,即以该节点为根节点的左右孩子不曾遍历. 当unUsed=0时,表示该节点的左右孩子都已经被访问过. 由于入栈的顺序和出栈的顺序相反,所以若unUsed=1,则左根右节点依次入栈,且根节…

[145-Binary Tree Postorder Traversal(二叉树非递归后序遍历)] [LeetCode-面试算法经典-Java实现][全部题目文件夹索引] 原题 Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [3,2,1]. Note: Recursive sol…