*/// 由于主函数的形式已经符合分治法需要的形式(具有合适的返回值),直接使用主函数做为递归函数vector<int>preorderTraversal(TreeNode * root){//递归三要素之定义// write your code herevector<int> result;if(root == nullptr) {returnresult;// 递归三要素之出口}// 获取左子树的遍历结果vector<int> ...
* Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: vector<int> preorderTraversal(TreeNode*root) { vector<int>result; stack<TreeNode *...
解法三 Morris Traversal 上边的两种解法,空间复杂度都是O(n),利用 Morris Traversal 可以使得空间复杂度变为 O(1)。 它的主要思想就是利用叶子节点的左右子树是 null ,所以我们可以利用这个空间去存我们需要的节点,详细的可以参考 94 题 中序遍历。 public List<Integer> preorderTraversal(TreeNode root) { Li...
classSolution {public: vector<int> preorderTraversal(TreeNode *root) { vector<int>res; stack<TreeNode *>s;if(root == NULL){//空树returnres; } s.push(root);//放树根while(!s.empty()){ TreeNode*cc =(); s.pop(); res.push_back(cc->val);//访问根if(cc->right !=NULL){ s.p...
Given a binary tree, return thepreordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 1. 2. 3. 4. 5. return[1,2,3]. Note:Recursive solution is trivial, could you do it iteratively?
Morris traversal: My code: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicList<Integer>preorderTraversal(TreeNoderoot){List<Integer>ret=newArrayList<...
[TreeNode]()varcurr=rootwhile(stack.nonEmpty||curr!=null){while(curr!=null){stack.push(curr)rs.append(curr.value)curr=curr.left}valnode=stack.pop()curr=node.right}rs.toList}defpreorderTraversalV4(root:TreeNode):List[Int]={if(root==null)returnNilvalrs=ListBuffer.empty[Int]valstack=Stack...
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...
题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 根据前序遍历和中序遍历结果构造二叉树。 思路分析: 分析二叉树前序遍历和中序遍历的结果我们发现: 二叉树前序遍历的第一个节点是根节点。 在中序遍历...
LeetCode:Binary Tree Preorder Traversal 题目:非递归实现二叉树的前序遍历。题目链接 算法1:使用栈的非递归遍历。先用根节点初始化栈,然后循环如下操作:访问栈顶节点,先后把栈顶节点右节点和左节点压栈(次序不能反,先右节点,后左节点),代码如下: 1/**2* Definition for binary tree3* struct TreeNode {4...