Binary Tree Level Order Traversal(二叉树的层次遍历) HoneyMoose iSharkFly - 鲨鱼君 来自专栏 · Java 描述给出一棵二叉树,返回其节点值的层次遍历(逐层从左往右访问)样例给一棵二叉树 {3,9,20,#,#,15,7}:3 / \ 9 20 / \ 15 7返回他的分层遍历结果:[ [3], [9,20], [15,7] ]挑战挑战1...
return its level order traversal as: [ [3], [9,20], [15,7] ] 代码: /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: ...
哦, 差点忘记提了, 还有一道关联题, https://leetcode.com/problems/binary-tree-level-order-traversal-ii/,要求改成了逆序输出, 即从 最后一层往前倒第一层地逐层遍历, 我想只要将结果集reverse下就好了吧, 或者一开始就递归求解?
push(c.left); if(c.right!=null) s2.push(c.right); } ans.add(tmp); tmp=new ArrayList<Integer>(); while(!s2.isEmpty()) { c=s2.pop(); tmp.add(c.val); if(c.right!=null)s1.push(c.right); if(c.left!=null)s1.push(c.left); } if(!tmp.isEmpty()) ans.add(tmp); }...
102 Binary Tree Level Order Traversal 用队列层次遍历 /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root). For example: Given binary tree [3,9,20,null,null,15,7], 3/\920/\157 1.
A C++ program that efficiently calculates the average values of nodes at each level in a binary tree, employing a level-order traversal approach for accurate and fast computation. queuecppbinary-treememory-managementtree-traversalcomputational-complexitylevel-order-traversalnode-averaging ...
Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its bottom-up level order traversal as: [[15,7],[9,20],[3]] 二叉树的层次遍历,不过需要保持每一层的数据,可以通过vector配合队列实现,实现: classSolution{public:vector<vector<int>>levelOrderBottom(TreeNode*root...
Given binary tree{3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3], [9,20], [15,7] ] 方法: 做了这么多层次遍历的变形了,终于做到原原本本的层次遍历了。写一下层次遍历的方法吧。使用queue数据结构;设立last指针,始终指向这一层的最后一个节点;设立...
{ root = NULL; //set root as NULL at the beginning } void levelorder_traversal(node *r); node *insert_node(node *root, int key); }; node *tree::get_node(int key){ node *new_node; new_node = new node; //create a new node dynamically new_node->h_left = 0; new_node->h...