1. Problem Descriptions:Given two integer arrays inorderandpostorderwhereinorderis the inorder traversal of a binary tree andpostorderis the postorder traversal of the same tree, construct and retu…
具体可见:https://leetcode.com/problems/binary-tree-inorder-traversal/solution/ /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: ...
Here is the definition for in-order traversal of a binary tree,click here. Time comlexity is O(n), space complexity is O(1), wherenis the number of nodes in the tree. Accepted code: 1/**2* Definition for binary tree3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode...
vector<int> inorderTraversal(TreeNode* root) { vector<int> res; stack<TreeNode*> treeroot; while(root!=NULL||!treeroot.empty()) { while(root!=NULL) { treeroot.push(root); root=root->left; } root = (); treeroot.pop(); res.push_back(root->val); root = root->right; } retu...
解法三 Morris Traversal 解法一和解法二本质上是一致的,都需要 O(h)的空间来保存上一层的信息。而我们注意到中序遍历,就是遍历完左子树,然后遍历根节点。如果我们把当前根节点存起来,然后遍历左子树,左子树遍历完以后回到当前根节点就可以了,怎么做到呢?
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 思路 这道题目,翻译一下就是:如何根据先序和中序遍历的数组,来构造二叉树。其中Note部分说,假设不存在重复项。为什么这样说呢?因为如果存在重复项,那么构造出来...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 翻译:给定树的前序和中序遍历,构造二叉树。 注意: 树中不存在重复项。 思路:首先,你应该知道 前序遍历:根节点,左子树,右子树; ...
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. 先序和中序、后序和中序可以唯一确定一棵二叉树 时间复杂度O(n),空间复杂度O(logn) // TreeNode.javapublicclassTreeNode{publicTreeNodeleft;publicTr...
Binary tree traversal: Preorder, Inorder, and Postorder In order to illustrate few of the binary tree traversals, let us consider the below binary tree: Preorder traversal: To traverse a binary tree in Preorder, following operations are carried-out (i) Visit the root, (ii) Traverse the le...
On the other hand, the cost of ordinary inorder traversal is linear, whether or not the tree is balanced. We present some data which suggests that the traversal time is O(n), and demonstrate an O(n log log n) upper bound. This upper bound is reached through a rather unusual unbounded...