We run a preorder depth first search on the rootof a binary tree. At each node in this traversal, we output D dashes (where D is the depth of this node), then we output the value of this node.  (If the depth of a node is D, the depth of its immediate…

题目如下: We run a preorder depth first search on the root of a binary tree. At each node in this traversal, we output D dashes (where D is the depth of this node), then we output the value of this node.  (If the depth of a node is D, the depth of its im…

We run a preorder depth first search on the root of a binary tree. At each node in this traversal, we output D dashes (where D is the depth of this node), then we output the value of this node.  (If the depth of a node is D, the depth of its immediat…

给定一个二叉树,以集合方式返回其中序/先序方式遍历的所有元素. 有两种方法,一种是经典的中序/先序方式的经典递归方式,另一种可以结合栈来实现非递归 Given a binary tree, return the inorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,3,2]. OJ's Binary Tree Serialization: The s…

思路:用一个栈来管理树的层次关系,索引代表节点的深度 方法一: TreeNode* recoverFromPreorder(string S) { /* 由题意知,最上层节点深度为0(数字前面0条横线),而第二层节点前有1条横线,表示深度为1 树的前序遍历: 根-左-右 因此, */ if (S.empty()) return nullptr; vector<TreeNode*> stack; // 结果栈 ,depth=,val=;i<S.size();) { ;i<S.size…

Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. 这道题要求用先序和中序遍历来建立二叉树,跟之前那道Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序遍历建立二叉树原理基本相同,针对这道题,由于先…

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 tree: 3 / \ 9 20…

Return any binary tree that matches the given preorder and postorder traversals. Values in the traversals pre and post are distinct positive integers. Example 1: Input: pre = [1,2,4,5,3,6,7], post = [4,5,2,6,7,3,1] Output: [1,2,3,4,5,6,7] Note: 1 <=…

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 = [,,,,] inorder = [,,,,] Return the following binary tree: / \ / \ 前序.中序遍历得到二叉树,可以知道…

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…