* Given a binary tree, return the inorder traversal of its nodes' values. * For example: * Given binary tree{1,#,2,3}, 1 \ 2 / 3 * return[1,3,2]. * Note: Recursive solution is trivial, could you do it iteratively?
classSolution(object):definorderTraversal(self, root):#递归""":type root: TreeNode :rtype: List[int]"""ifnotroot:return[]returnself.inorderTraversal(root.left) + [root.val] + self.inorderTraversal(root.right) Java解法 classSolution {publicList < Integer >inorderTraversal(TreeNode root) ...
6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<int> inorderTraversal(TreeNode *root) { 13 vector<int> res; 14 if(root==NULL) return res; 15 stack<TreeNode*> s; 16 TreeNode* cur=...
public: vector<int> inorderTraversal(TreeNode* root) { vector<int> res; stack<TreeNode*> treeroot; while(root!=NULL||!treeroot.empty()) { while(root!=NULL) { treeroot.push(root); root=root->left; } root = (); treeroot.pop(); res.push_back(root->val); root = root->right;...
Given a binary tree, return the inorder traversal of its nodes' values. 给定一个二叉树,返回中序遍历后所有节点的值。 Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2] Follow up: Recursive solution is trivial, could you do it iteratively? 追加问题:可以不用递归,而是用遍历实...
public List<Integer> inorderTraversal(TreeNode root) { List<Integer> ans = new ArrayList<>(); getAns(root, ans); return ans; } private void getAns(TreeNode node, List<Integer> ans) { if (node == null) { return; } getAns(node.left, ans); ans.add(node.val); getAns(node.right...
*/classSolution{public List<Integer>preorderTraversal(TreeNode root){List<Integer>result=newLinkedList<>();TreeNode current=root;TreeNode prev=null;while(current!=null){if(current.left==null){result.add(current.val);current=current.right;}else{// has left, then find the rightmost of left su...
名字叫做, morrois traversal, 自己写了下: My code: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicList<Integer>inorderTraversal(TreeNoderoot){List...
Splaying a search tree in preorder takes linear time - Chaudhuri, Hoft - 1993 () Citation Context ... any O(1)-competitive binary search tree must have the traversal property, no binary search tree is known to have the traversal property. However, special case of the traversal property (...
Given a binary tree, return theinordertraversal of its nodes' values. 示例: 代码语言:javascript 代码运行次数:0 AI代码解释 输入:[1,null,2,3]1\2/3输出:[1,3,2] 进阶:递归算法很简单,你可以通过迭代算法完成吗? Follow up:Recursive solution is trivial, could you do it iteratively?