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->...
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: vector<int> preorderTraversal(TreeNode*root) { vector<int>result; preorder(root, result);returnresult; }voidpreorder(TreeNode *root, vector<int> &result) {if(root !=nullptr) { result.push_back...
classSolution {public: vector<int> preorderTraversal(TreeNode *root) { vector<int>res; stack<TreeNode *>s;if(root == NULL){//空树returnres; } s.push(root);//放树根while(!s.empty()){ TreeNode*cc =(); s.pop(); res.push_back(cc->val);//访问根if(cc->right !=NULL){ s.p...
上边的两种解法,空间复杂度都是O(n),利用 Morris Traversal 可以使得空间复杂度变为O(1)。 它的主要思想就是利用叶子节点的左右子树是null,所以我们可以利用这个空间去存我们需要的节点,详细的可以参考94 题中序遍历。 publicList<Integer>preorderTraversal(TreeNoderoot){List<Integer>list=newArrayList<>();TreeN...
144. Binary Tree Preorder Traversal刷题笔记 问题描述前序遍历。注意嵌套函数里的list应该用append而不是+=来添加元素 # Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val...
printPreorder(root.left)printPreorder(root.right)# Driver coderoot=Node(1)root.left=Node(2)root.right=Node(3)root.left.left=Node(4)root.left.right=Node(5)print("Preorder traversal of binary tree is")printPreorder(root)print("\nInorder traversal of binary tree is")printInorder(root)...
INORDER AND PREORDER TRAVERSAL and DISPLAY OF EXISTING BINARY TREE IN COMPUTER MEMORYP K KumaresanJournal of Harmonized Research
15、课程:树(下).10、练习—Iterative Postorder Traversal -- -- 10:33 App 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...
* TreeNode(int x) { val = x; } * } */classSolution{public List<Integer>inorderTraversal(TreeNode root){List<Integer>result=newLinkedList<>();TreeNode current=root;TreeNode prev=null;while(current!=null){// left firstif(current.left==null){result.add(current.val);current=current.right...
Given a binary tree, return the preorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 2 / 3 Output: [1,2,3] Follow up: Recursive solution is trivial, could you do it iteratively? 翻译 给出一棵二叉树,返回其节点值的前序遍历。