new_node->right = NULL; return new_node; } void tree::levelorder_traversal(node *root){ queue <node*> que; node *item; que.push(root); //insert the root at first while(!que.empty()){ item = que.front(); //get the element from the front end cout << item->value << " "...
题目Given a binary tree, return the inorder traversal of its nodes' values. For example: Given binary tree [1,null,2,3], return [1,3,2]. Note: Recursive solution is trivial, could you do it iter…
*right;7* TreeNode(int val) {8* this->val = val;9* this->left = this->right = NULL;10* }11* }12*/13classSolution {14/**15* @param root: The root of binary tree.16* @return: Inorder in vector which contains node values.17*/18public:19vector<int> inorderTraversal...
A binary tree isthreadedby making all right child pointers that would normally be null point to the inorder successor of the node (ifit exists), and all left child pointers that would normally be null point to the inorder predecessor of the node. 就是说线索二叉树实际上是把所有原本为空的...
push(root); 28 tree_queue.push(NULL); 29 int nLevelCount = 1; 30 while (true) { 31 TreeNode *pTemp = tree_queue.front(); 32 tree_queue.pop(); 33 if (pTemp == NULL) { 34 if (nLevelCount%2 == 0) { //if the num of level is odd, swap the ivec; 35 Swap(ivec); ...
binary treesdata arrayauxiliary stackpointer reversal methodPASCAL/ C6120 File organisation C6140D High level languagesA non-recursive algorithm for the traversal of a binary tree is presented in which the order of traversal is defined by an external data array, allowing any of the six possible ...
Given a binary tree, return theinordertraversal of its nodes' values. 示例: 代码语言:javascript 代码运行次数:0 AI代码解释 输入:[1,null,2,3]1\2/3输出:[1,3,2] 进阶:递归算法很简单,你可以通过迭代算法完成吗? Follow up:Recursive solution is trivial, could you do it iteratively?
18 A and B are two nodes on a binary tree. In the in-order traversal, the condition for A before B is ( ). A. A is to the left of B B. A is to the right of B C. A is the ancestor of B D. A is a descendant of B ...
// Binary Tree Level Order Traversal II class Solution_107 { public: //bfs vector<vector<int>> levelOrderBottom(TreeNode* root) { vector<vector<int> > vecs; stack<vector<int>> sta; if (!root) { return vecs; } queue<TreeNode*> que; ...
105. Construct Binary Tree from Preorder and Inorder Traversal,这道题用指针,分治思想,相信看代码就能很容易弄懂了这里有一个问题未解决(希望有人可以回答一下:buildTree函数如果不加if语句在input为两个空vector对象的时候报错,搞不清楚为什么,因为我的build函