Given a binary search tree, print the elements in-order iteratively without using recursion.Note:Before you attempt this problem, you might want to try coding a pre-order traversal iterative solution first, because it is easier. On the other hand, coding a post-order iterative version is a ...
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) ...
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; ...
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 ...
*/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...
Binary tree traversal: Preorder, Inorder, and Postorder In order to illustrate few of the binary tree traversals, let us consider the below binary tree: Preorder traversal: To traverse a binary tree in Preorder, following operations are carried-out (i) Visit the root, (ii) Traverse the le...
Java Solution 1 – Iterative The key to solve inorder traversal of binary tree includes the following: The order of “inorder” is: left child -> parent -> right child Use a stack to track nodes publicList<Integer>inorderTraversal(TreeNoderoot){ArrayList<Integer>result=newArrayList<>();Stac...
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]. 栈迭代 复杂度 时间O(b^(h+1)-1) 空间 O(h) 递归栈空间 对于二叉树b=2 ...
publicList<Integer>inorderTraversal3(TreeNoderoot){List<Integer>ans=newArrayList<>();TreeNodecur=root;while(cur!=null){//情况 1if(cur.left==null){ans.add(cur.val);cur=cur.right;}else{//找左子树最右边的节点TreeNodepre=cur.left;while(pre.right!=null&&pre.right!=cur){pre=pre.right;}...
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...