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();15rVec.push_back(tree->val);16st.pop();17tree = tree->...
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...
0,preorder.length,inorder,0,inorder.length);}privateTreeNodebuildTreeHelper(int[]preorder,intp_start,intp_end,int[]inorder,inti_start,inti_end){// preorder 为空,直接返回 nullif(p_start==p_end){returnnull;}introot_val=preorder[p_start];TreeNoderoot=newTreeNode(root_val);//在中序遍...
usingSystem.Text; namespacebinarytree { #region 节点的定义 classnode { publicstringnodevalue; publicnode leftchild, rightchild; publicnode() { } publicnode(stringvalue) { nodevalue = value; } publicvoidassignchild(node left, node right)//设定左右孩子 { this.leftchild = left; this.rightchild...
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...
order B. Traversing the forest corresponding to the binary tree in root-last order C. Traversing the forest corresponding to the binary tree in breadth-first order D. None of the above 答案：A 分析：正确答案:A 解析:前序遍历一个二叉树等价于按从树的根部、右子树、右子树查找顺序查找树。
Binary treeIn most of the Data Structure textbooks there is a conclusion: "Given the preoder and the inorder sequence of nodes, the binary tree structure can be determined exclusively". This paper gives a counter example to prove this conclusion is wrong, and analzes the reason which cause...
15、课程:树(下).7、练习—Iterative Get和Iterative Add 2 -- 12:36 App 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...
Memory Usage: 13.1 MB, less than 70.72% of Python3 online submissions for Binary Tree Inorder Traversal. None-trivial solution: NOTE: Maintain a stack, push root first, then every iteration, push left child first if existed. If a left child is searched, result.append(current_parent_node's...
Given binary tree{1,#,2,3}, 1 \ 2 / 3 return[1,2,3]. 和上面要求一样,只是要返回以中序方式序列的元素,这次用递归实现: 1/**2* Definition for binary tree3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x), left(NULL), right...