int BST_Insert(BSTree &T, int k, Node* parent=NULL) { if(T == NULL) { T = (BSTree)malloc(sizeof(Node)); T->key = k; T->left = NULL; T->right = NULL; T->parent = parent; return 1; // 返回1表示成功 }else if(...
link:[https://leetcode.com/explore/learn/card/data-structure-tree/134/traverse-a-tree/928/] 递归解法: #Definition for a binary tree node.#class TreeNode(object):#def __init__(self, x):#self.val = x#self.left = None#self.right = NoneclassSolution(object):defsolve(self,root):ifroo...
class Solution { public int widthOfBinaryTree(TreeNode root) { int result = 0; Deque<TreeNode> deque = new ArrayDeque(); deque.addLast(new TreeNode(1, root.left, root.right)); while(!deque.isEmpty()) { int count = deque.size(), startIndex = -1, endIndex = -1; for (int i ...
int*returnSize){//每次调用函数时,都要把i初始化//如果没有初始化,则i会一直叠加,无法重复使用i=0;// 调用TreeSize函数计算二叉树的节点数int size=TreeSize(root);// 动态分配结果数组,大小为节点数int*a=(int*)malloc(size*sizeof(int));// 调用辅助函数_prevOrder执行前序遍历,填充数组a_prevOrder(...
/** * Definition for a binary tree node. * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */ struct TreeNode* insertIntoBST(struct TreeNode* root, int val) { struct TreeNode* node = (struct TreeNode*)(malloc(sizeof(struct TreeNode)))...
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right (right) {} * }; */ classSolution{ public: intwidthOfBinaryTree(TreeNode*root) { ...
(ans,0,root)returnans}funvisitLevel(ans:MutableList<MutableList<Int>>,level:Int,node:TreeNode?){if(null==node)return// level 从0 (root) 开始,此时 ans.size = 0; 每层的值存在 levelList 中. 这地方的代码非常巧妙.if(ans.size==level){val levelList=mutableListOf<Int>()ans.add(levelList...
The size of the given array will be in the range [1,1000]. 题意:给定一个一维数组,里面的元素代表树的节点;现在要求构造一颗maximum tree;具体要求如下 根节点的值为数组中最大的那个数 左子树的根节点的值为最大数所在数组位置左半部分数组中的最大值 ...
You are given a binary tree in which each node contains an integer value.Find the number of paths that sum to a given value.The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes). The tree has no...
public int widthOfBinaryTree(TreeNode root) { int[] ans = new int[1]; robot(root, ans, new ArrayList<>(), 0, 1); return ans[0]; } private void robot(TreeNode root, int[] ans, ArrayList<Integer> leftIndexes, int level, int index) { if (root == null) { return; } if (le...