because it is easier. On the other hand, coding a post-order iterative version is a challenge. See my post:Binary Tree Post-Order Traversal Iterative Solutionfor more details and an in-depth analysis of the pro
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) ...
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 ...
Leetcode 94: Binary Tree Inorder Traversal-中序遍历 自动驾驶搬砖师 一步一个台阶,一天一个脚印 一、题目描述 给定一个二叉树,返回它的中序遍历。 二、解题思路 2.1 递归法 时间复杂度 O(n) 空间复杂度 O(n) 2.2 迭代法 时间复杂度 O(n) 空间复杂度 O(n)(1) 同理创建一个Stack,然后按左/中/右...
self.inorder(root.right, list) def inorderTraversal(self, root): """ :type root: TreeNode :rtype: List[int] """ if not root: return [] result = [] self.generate(root, result) return result 1. 2. 3. 4. 5. 6. 7.
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; ...
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
*/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;}// if there is left, get the rightmost...
从大的层面讲,Binary Tree 可以用DFS和BFS。 对于BFS,我们需要iterative with queue。 对于DFS,Binary Tree 有三种traversal的方式: ** Inorder Traversal: left -> root -> right ** Preoder Traversal: root -> left -> right ** Postoder Traveral: left -> right -> root ...
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...