*/// 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> ...
题目地址:https://leetcode.com/problems/binary-tree-preorder-traversal/description/ Given a binary tree, return thepreordertraversal of its nodes' values. Example: Input:[1,null,2,3]1 \ 2 / 3Output:[1,2,3] Follow up:Recursive solution is trivial, could you do it iteratively? 给定一个...
Given a binary tree, return the preorder traversal of its nodes' values. For example: Given binary tree {1,#,2,3}, 1 \ 2 / 3 return [1,2,3]. Note: Recursive solution is trivial, could you do it iteratively? SOLUTION1&2: 递归及非递归解法: View Code https://github.com/yuzhang...
*/voidtraversal(structTreeNode*root,int*index,int*res){if(!root)return;res[(*index)++]=root->val;traversal(root->left,index,res);traversal(root->right,index,res);}int*preorderTraversal(structTreeNode*root,int*returnSize){int*res=malloc(sizeof(int)*110);intindex=0;traversal(root,&index,...
Binary Tree Preorder Traversal 题目链接 题目要求: 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?
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。 publicList<Integer>preorderTraversal(TreeNoderoot){List<Integer>list=newArrayList<>();Tree...
每天一算:Binary Tree Preorder Traversal 程序员吴师兄 + 关注 预计阅读时间2分钟 6 年前 LeetCode上第144 号问题:二叉树的前序遍历 题目 给定一个二叉树,返回它的 前序 遍历。 示例: 输入: [1,null,2,3] 输出: [1,2,3] 进阶: 递归算法很简单,你可以通过迭代算法完成吗? 解题思路 用栈(Stack)...
{public:vector<int>preorderTraversal(TreeNode*root){vector<int>result;stack<TreeNode*>st;if(root==NULL)returnresult;st.push(root);while(!st.empty()){TreeNode*top=st.top();st.pop();result.push_back(top->val);if(top->right)st.push(top->right);if(top->left)st.push(top->left);...
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<int> preorderTraversal(TreeNode* root) {...
self.val = val self.left = left self.right = right class Solution: # 递归 def preorderTraversal_1(self, root: TreeNode) -> List[int]: res = [] self.dfs(root, res) return res def dfs(self, node, res): if not node: return ...