1vector<int> inorderTraversal(TreeNode*root) {2vector<int>rVec;3stack<TreeNode *>st;4TreeNode *tree =root;5while(tree || !st.empty())6{7if(tree)8{9st.push(tree);10tree = tree->left;11}12else13{14tree =st.top();
但是inorder不简单啊,模仿递归记录调用处然后处理完当前函数回来,下午困得不行闷头试了好几次各种bug超Memory,看了我是歌手回来发现脑子清醒了 1vector<int> inorderTraversal(TreeNode *root) {2vector<int>nodes;3if(root == NULL)returnnodes;4stack<TreeNode *>tStack;5TreeNode * cur =root;6while(!tS...
publicTreeNodebuildTree(int[]preorder,int[]inorder){HashMap<Integer,Integer>map=newHashMap<>();for(inti=0;i<inorder.length;i++){map.put(inorder[i],i);}returnbuildTreeHelper(preorder,0,preorder.length,inorder,0,inorder.length,map);}privateTreeNodebuildTreeHelper(int[]preorder,intp_st...
那就是在inorder获得的左子树的长度+当前的index(inorder的作用就体现出来了,因为只有他存在,我们才知道每个根节点的左右子树长度,才知道哪里停止)。 3.进入到下一个节点时,我们对inorder也要进行分割来获得左子树的inorder的排列。 int[] left_tree_in = inorder[:curr_root_index_in_inorder]; 这么做的理...
inorder_visit(node_a);//中序 Console.WriteLine(); postorder_visit(node_a);//后序 Console.WriteLine(); node node_1 =newnode("1"); node node_2 =newnode("2"); node node_3 =newnode("3"); node node_4 =newnode("4");
LeetCode: 105. Construct Binary Tree from Preorder and Inorder Traversal 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. For example, given ...
INORDER AND PREORDER TRAVERSAL and DISPLAY OF EXISTING BINARY TREE IN COMPUTER MEMORYP K KumaresanJournal of Harmonized Research
15、课程:树(下).13、练习—Construct Binary Tree from Preorder and Inorder Traversal 1 -- 9:07 App 15、课程:树(下).3、练习—Floor and Ceiling 11 -- 12:45 App 13、课程:哈希表(下).10、作业讲解 3 -- 6:45 App 10、课程:堆栈和队列(上).10、练习—Min stack(一) 2 -- 6:07...
树的三种DFS遍历,是指按照根节点(自己)被访问的顺序 Pre-Order: 先访问根节点,再访问其左右子树。对每一个subtree,同样适用。 In-Order: 先访问其...
Given a binary tree, return the inorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 2 / 3 Output: [1,3,2] Follow up: Recursive solution is trivial, could you do it iteratively? Solutions: Trivial solution: ...