* 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: vector<int> inorderTraversal(TreeNode*root) { vector<int>result; helper(root, resul...
*/voidinorder(structTreeNode* root,int* res,int* resSize){if(!root) {return; } inorder(root->left, res, resSize); res[(*resSize)++] = root->val; inorder(root->right, res, resSize); }int*inorderTraversal(structTreeNode* root,int* returnSize){int* res =malloc(sizeof(int) *...
Given the root of a binary tree, return the inorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [1,3,2] Example 2: Input: root = [] Output: [] Example 3: Input: root = [1] Output: [1] Constraints: The number of nodes in the tree is...
publicList<Integer>inorderTraversal(TreeNoderoot){List<Integer>ans=newArrayList<>();getAns(root,ans);returnans;}privatevoidgetAns(TreeNodenode,List<Integer>ans){if(node==null){return;}getAns(node.left,ans);ans.add(node.val);getAns(node.right,ans);} 时间复杂度:O(n),遍历每个节点。 空间...
* TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: /*递归版 vector<int> inorderTraversal(TreeNode *root) { vector<int> res; solve(root,res); return res; } void solve(TreeNode *root,vector<int> &res) ...
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution{public:vector<int>inorderTraversal(TreeNode*root){vector<int>res;inorder(root,res);returnres;}voidinorder(TreeNode*root,vector<int>&res){if(!root)return;if(root->left)inorder(root->left,res);res.pu...
* 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 thepreordertraversal of its nodes' values. Example: Input:[1,null,2,3]1 \ 2 / 3Output:[1,2,3] Follow up:Recursive solution is trivial, could you do it iteratively? 给定一个二叉树,返回它的前序遍历。
In this article, we will discuss 3 different techniques for Level Order Traversal of a binary tree. This technique can be used to find the left and right view of the tree.
val right=node.rightif(null!=left)visitLevel(ans,level+1,left)if(null!=right)visitLevel(ans,level+1,right)}classTreeNode(var`val`:Int){varleft:TreeNode?=nullvarright:TreeNode?=null} 参考资料 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/binary-tree-level-order-traversal著作...