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 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 ...
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; } retu...
public List<Integer> inorderTraversal3(TreeNode root) { List<Integer> ans = new ArrayList<>(); TreeNode cur = root; while (cur != null) { //情况 1 if (cur.left == null) { ans.add(cur.val); cur = cur.right; } else { //找左子树最右边的节点 TreeNode pre = cur.left; whi...
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...
94. Binary Tree Inorder Traversal 两种解法,递归和迭代 Memory分别是6.1和5.7,速度好像是一样的。。 Recursive class Solution { public: vector<int>res; void in_order(TreeNode* root) { if (!root)return; in_order(root->left); res.push_back(root->val);...
二叉树层序遍历 Example Givenbinary tree[3,9,20,null,null,15,7],3/\920/\157returnits level order traversalas:[[3],[9,20],[15,7]] BFS方法 var levelOrder=function(root){if(!root)return[]conststack=[root]constres=[]while(stack.length){constlen=stack.length;// 每一层遍历完,会重新...
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 翻译:给定树的前序和中序遍历,构造二叉树。 注意: 树中不存在重复项。 思路:首先,你应该知道 前序遍历:根节点,左子树,右子树; ...
Preorder traversal: To traverse a binary tree in Preorder, following operations are carried-out (i) Visit the root, (ii) Traverse the left subtree, and (iii) Traverse the right subtree. Therefore, the Preorder traversal of the above tree will outputs: ...
Can we build the tree only from postorder traversal?Of course no. Because, though we find the root as the last element in the postorder traversal, we can't find its subtrees from the rest of the traversal.For example, say the postorder traversal is:...