* @param root: A Tree * @return: Inorder in ArrayList which contains node values. */vector<int>inorderTraversal(TreeNode * root){// write your code herevector<int> result; traverse(root, result);returnresult; }/
binary-tree-inorder-traversal /** * * @author gentleKay * 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?
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...
2.2 last.right 不为 null,说明之前已经访问过,第二次来到这里,表明当前子树遍历完成,保存 cur 的值,更新 cur = cur.right public List<Integer> inorderTraversal3(TreeNode root) { List<Integer> ans = new ArrayList<>(); TreeNode cur = root; while (cur != null) { //情况 1 if (cur.left ...
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?
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; ...
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example: Given binary tree [3,9,20,null,null,15,7], 代码语言:javascript 代码运行次数:0 运行 AI代码解释 ...
Using these new definitions, the leaf nodes in binary tree (a) are nodes 6 and 8; the internal nodes are nodes 1, 2, 3, 4, 5, and 7.Unfortunately, the .NET Framework does not contain a binary tree class, so in order to better understand binary trees, let's take a moment to ...
Can you solve this real interview question? Binary Tree Preorder Traversal - Given the root of a binary tree, return the preorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [1,2,3] Explanation: [https://assets.l
Similar to pre-order and in-order traversal, accessing the binary tree in the order of left child->right child->root node is called post-order traversal. The first visited node in the post-order traversal is与先序、中序遍历类似,以左子->右子->根节点的顺序来访问二叉树称为后序遍历。后序...