class BSTIterator { Queue<Integer> queue = new LinkedList<>(); public BSTIterator(TreeNode root) { inorderTraversal(root); } private void inorderTraversal(TreeNode root) { if (root == null) { return; } inorderT
Val int * Left *TreeNode * Right *TreeNode * } */ type BSTIterator struct { // 结点栈(所有结点的左子结点都已经入栈过) stack []*TreeNode } func Constructor(root *TreeNode) BSTIterator { // 初始化一个空栈的迭代器 bstIterator := BSTIterator{} // 将 root 放入迭代器中 bstIterator...
最简单的,先把整棵树给遍历了,找到所有的值 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(NULL) {}8* };9*/10classBSTIterator {11public:1213vector<int>vals;14intindex;15B...
Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. Callingnext()will return the next smallest number in the BST. Note:next()andhasNext()should run in average O(1) time and uses O(h) memory, wherehis the height of t...
Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. Calling next() will return the next smallest number in the BST. Note: next() and hasNext() should run in average O(1) ti...
* TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class BSTIterator { public: stack<TreeNode *> ss; // 初始化时,栈里保存从根节点到最左边叶子的节点,栈顶是整颗树的最小值 //从根向左遍历 最小值在栈顶 ...
祖先(ancestor)结点无序树 子孙(descendant)结点森林 结点所处层次(level)树的高度(depth)树的度(degree)树的抽象数据类型 template<classType>classTree{public:Tree();~Tree();positionRoot();BuildRoot(constType&value);positionFirstChild(positionp);positionNextSibling(positionp,positionv);positionParent(...
Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. 栈迭代 复杂度 时间O(b^(h+1)-1) 空间 O(h) 递归栈空间 对于二叉树b=2
* TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassBSTIterator{Stack<TreeNode>s;publicBSTIterator(TreeNoderoot){s=newStack<TreeNode>();while(root!=null){s.push(root);root=root.left;}}/** @return whether we have a next smallest number */publicbooleanhasNext(){re...
You can return the inorder traversal of a BST tree [1,2,3] Challenge Extra memory usage O(h), h is the height of the tree. Super Star: Extra memory usage O(1) 思路: 该Iterator 算法即 non-recursion 的 inorder traversal,不仅仅适用于 BST,任何 Binary Tree 都可以 ...