1的left - preOrder中1的下一个数字2, 范围缩小到inOrder position [0, 2],2在其中,设2为1的left, 递归2; 2的left - preOrder中2的下一个数字3, 范围缩小到inOrder position [0, 0],3在其中,设3为2的left, 递归3; 3的left - preOrder中3的下一个数字4, 范围缩小到inOrder position [0, -...
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->...
Given preorder and inorder traversal of a tree, construct the binary tree. **Note:**You may assume that duplicates do not exist in the tree. For example, given preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] Return the following binary tree: 3 / \ 9 20 / \ 15 7 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. recursive answer: 递归方法:在preorder找到root节点,接着在inorder里找到root节点,则inorder被分为两部分,一部分是left,一部分是right。在确定right支的时候要注...
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; ...
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:
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...
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。 publicList<Integer>preorderTraversal(TreeNoderoot){List<Integer>list...
a binary tree with a pre-order traversal algorithm using Recursion, and in this article, you will learn how to implement pre-order traversalwithout using Recursion. You might be thinking that why do you need to learn the iterative solution if a recursive solution is possible and easy to ...
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: ...