Given therootof a binary search tree, returna balanced binary search tree with the same node values. If there is more than one answer, return any of them. A binary search tree is balanced if the depth of the two subtrees of every node never differs by more than1. Example 1: Input: ...
Can you solve this real interview question? Balance a Binary Search Tree - Given the root of a binary search tree, return a balanced binary search tree with the same node values. If there is more than one answer, return any of them. A binary search tree
TreeNode rightChild;intvalue;publicTreeNode(intvalue){this.value=value; }publicTreeNode(){ } }publicstaticvoidmain(String[] args){ BalancedBinaryTree balancedBinaryTree =newBalancedBinaryTree();char[] arr0 =newchar[]{'#'};char[] arr1 =newchar[]{'3','9','2','#','#','1','7'}...
构造二叉树 root := &TreeNode{Val: 1} root.Left = &TreeNode{Val: 2} root.Rig...
/** * Definition for a binary tree node. * type TreeNode struct { * 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...
Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as: a binary tree in which the depth of the two subtrees of every node never differ by more than 1. 1. Example 1: ...
Can you solve this real interview question? Binary Search Tree Iterator - Implement the BSTIterator class that represents an iterator over the in-order traversal [https://en.wikipedia.org/wiki/Tree_traversal#In-order_(LNR)] of a binary search tree (BST):
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees ofeverynode never differ by more than 1. 保证左右子树的深度差不超过1 classSolution {public:intdepth(TreeNode *root){if(!root)return0;return1+max(depth(root->left),depth...
第C++实现LeetCode(108.将有序数组转为二叉搜索树)[LeetCode]108.ConvertSortedArraytoBinarySearchTree将有序数组转为二叉搜索树 Givenanarraywhereelementsaresortedinascendingorder,convertittoaheightbalancedBST. Forthisproblem,aheight-balancedbinarytreeisdefinedasabinarytreeinwhichthedepthofthetwosubtreesofeverynode...
TreeNode* build(int l, int r) { int mid = (l + r) >> 1; TreeNode* o = new TreeNode(inorderSeq[mid]); if (l <= mid - 1) { o->left = build(l, mid - 1); } if (mid + 1 <= r) { o->right = build(mid + 1, r); ...