https://leetcode.com/problems/binary-tree-postorder-traversal/description/ Description Implement an iterative, post-order traversal of a given binary tree, return the list of keys of each node in the tree as it
voidpostOrderTraversal(BinaryTree *p) {if(!p)return; postOrderTraversal(p->left); postOrderTraversal(p->right); cout<< p->data; } This is the most difficult of all types of iterative tree traversals. You should attempt this problem:Binary Search Tree In-Order Traversal Iterative Solutionbef...
If I understand the exercise correctly, you would have to come up with formulas for how to calculate the label of the left and right child during each method of traversal, given the label of the current node and the depth. For example, during pre-order traversal, the lab...
The pre-order and post-order traversal of a Binary Tree generates the same output. The tree can have maximum . A、Three nodes B、Two nodes C、One node D、Any number of nodes 暂无答案
The pre-order and post-order traversal of a Binary Tree is A 、B 、C and C 、B 、A, respectively. The number of trees that satisfy the conditions is . A.1 B.2 C.3 D.4 暂无答案
*/classSolution{public List<Integer>preorderTraversal(TreeNode root){List<Integer>result=newLinkedList<>();TreeNode current=root;TreeNode prev=null;while(current!=null){if(current.left==null){result.add(current.val);current=current.right;}else{// has left, then find the rightmost of left su...
classSolution{voidpushLeft(TreeNode*root,stack<TreeNode*>&nodes){// this function here push all the LEFT node we can get from the input nodewhile(root){nodes.emplace(root);// emplace c++11root=root->left;}}public:vector<int>inorderTraversal(TreeNode*root){vector<int>res;if(!root)return...
Given a binary tree, return thepostordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 return[3,2,1]. Note: Recursive solution is trivial, could you do it iteratively? 代码: stack 1:
postorderTraversal(result, root.left); postorderTraversal(result, root.right); result.add(root.val); } public void inorderTraversal(List<Integer> result, TreeNode root){ if(root==null) return; inorderTraversal(result, root.left); result.add(root.val); ...
Similar to pre-order and in-order traversal, accessing the binary tree in the order of left child->right child->root node is called post-order traversal. The first visited node in the post-order traversal is与先序、中序遍历类似,以左子->右子->根节点的顺序来访问二叉树称为后序遍历。后序...