class Solution { public: vector<int> inorderTraversal(TreeNode *root) { // Start typing your C/C++ solution below // DO NOT write int main() function vector<int> *p_vec = new vector<int>(); inorderRecur( root, *p_vec); return *p_vec; } void inorderRecur( TreeNode *root, vecto...
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: vector<int> inorderTraversal(TreeNode *root) { vector<int>ret;if(!root)returnret; inorder(root, ret);returnret; }voidinorder(TreeNode* root, vector<int>&ret) {if(root->left) inorder(root-...
Binary Tree Inorder Traversal 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? confused what "{1,#,2,3}" means? > ...
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) ...
* 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?
【Inorder Traversal】 Given a binary tree, return theinordertraversal 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?
Given a binary tree, return theinordertraversal 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? Analysis: 给出一棵二叉树,返回它节点值的中序遍历。
下面这种解法跟Binary Tree Preorder Traversal中的解法二几乎一样,就是把结点值加入结果 res 的步骤从 if 中移动到了 else 中,因为中序遍历的顺序是左-根-右,参见代码如下: 解法三: classSolution {public: vector<int> inorderTraversal(TreeNode*root) { ...
Solution classSolution{private:voidrecursiveBuild(TreeNode* &node, vector<int>& preorder,intpstart,intpend, vector<int>& inorder,intistart,intiend){if(pstart > pend)return; node =newTreeNode(preorder[pstart]);if(pstart == pend)return;inti;for(i = istart; i <= iend; i++)if(inord...
Given a binary tree, return theinordertraversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [1,3,2] Follow up: Recursive solution is trivial, could you do it iteratively? 回到顶部 Intuition 1. Recursion ...