*/// 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> ...
1vector<int> inorderTraversal(TreeNode*root) {2vector<int>rVec;3stack<TreeNode *>st;4TreeNode *tree =root;5while(tree || !st.empty())6{7if(tree)8{9st.push(tree);10tree = tree->left;11}12else13{14tree =st.top();15rVec.push_back(tree->val);16st.pop();17tree = tree->...
*/publicclassSolution{public List<Integer>preorderTraversal(TreeNode root){Stack<TreeNode>stack=newStack<>();List<Integer>result=newArrayList<>();if(root==null){returnresult;}stack.push(root);while(!stack.isEmpty()){TreeNode node=stack.pop();result.add(node.val);// 关键点:要先压入右孩...
1/**2* Definition for a binary tree node.3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x), left(NULL), right(NULL) {}8* };9*/10classSolution {11public:12vector<int> preorderTraversal(TreeNode*root) {13vector<int>rVec;14pr...
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? 给定一个二叉树,返回它的前序遍历。
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...
Given therootof a binary tree, returnthe preorder traversal of its nodes' values. Example 1: image <pre>Input:root = [1,null,2,3] Output:[1,2,3] </pre> Example 2: <pre>Input:root = [] Output:[] </pre> Example 3:
同Construct Binary Tree from Inorder and Postorder Traversal(Python版) # Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): # self.val = val # self.left = left # self.right = right class Solution: def buildTree(self, pre...
15、课程:树(下).10、练习—Iterative Postorder Traversal -- -- 10:33 App 15、课程:树(下).7、练习—Iterative Get和Iterative Add 2 -- 12:36 App 15、课程:树(下).13、练习—Construct Binary Tree from Preorder and Inorder Traversal 1 -- 9:07 App 15、课程:树(下).3、练习—Floor and...
/** * * @author gentleKay * Given a binary tree, return the preorder traversal of its nodes' values. * For example: * Given binary tree{1,#,2,3},