1.因为二叉搜索树的特性,将preorder数组排序,得到inorder。再将inorder的元素和下标用map存储起来,再对其进行递归。 2.利用二叉树的特性,初始化最小值,最大值,进行递归 3.用栈结构进行迭代。 classSolution {int[] preorder;intidx = 0; Map<Integer, Integer> map_inorder =newHashMap<>();publicTreeNode...
node->left = dfs(l +1, i -1, p); node->right =dfs(i, r, p);returnnode; } TreeNode* bstFromPreorder(vector<int>&preorder) {returndfs(0, preorder.size() -1, preorder); } }; 2. O(N) classSolution {public:inti =0; TreeNode* bstFromPreorder(vector<int>& A,intbound =...
current->right=build(p_l+left_tree_n+1,p_r,k+1,i_r); return current; } TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) { if(preorder.empty())return NULL; return build(preorder.begin(),preorder.end()-1,inorder.begin(),inorder.end()-1); } }; 1. 2. ...
TreeNode getTree(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd) { if (preStart >= preEnd) { return null; } int cur = preorder[preStart]; TreeNode root = new TreeNode(cur); int i = 0; int j = inStart; while(inorder[j] != cur) { ...
public List<Integer> preorderTraversal(TreeNode root) { List<Integer> list = new ArrayList<>(); preorderTraversalHelper(root, list); return list; } private void preorderTraversalHelper(TreeNode root, List<Integer> list) { if (root == null) { return; } list.add(root.val); preorderTrav...
Construct Binary Tree from Preorder and Inorder Traversal 题目描述(中等难度) 根据二叉树的先序遍历和中序遍历还原二叉树。 解法一 递归 先序遍历的顺序是根节点,左子树,右子树。中序遍历的顺序是左子树,根节点,右子树。 所以我们只需要根据先序遍历得到根节点,然后在中序遍历中找到根节点的位置,它的左边就...
Preorder traversal Visit root node Visit all the nodes in the left subtree Visit all the nodes in the right subtree display(root->data) preorder(root->left) preorder(root->right) Postorder traversal Visit all the nodes in the left subtree Visit all the nodes in the right subtree Visit th...
Learn how to perform pre-order traversal in a JavaScript tree and understand the process with examples and explanations.
用preorder traversal做一下就好了。 其实比较routine classSolution{publicTreeNoderecoverFromPreorder(StringS){int[]index=newint[]{-1};returnhelper(S,index,0);}privateTreeNodehelper(Strings,int[]index,intdepth){intfast=index[0];if(fast+1>=s.length())returnnull;while(fast+1<s.length()&&s....
preorder 的范围 [begin, end] 并不是 inorder 的范围,如果直接拿begin, end 来作为inorder[] 数组的边界条件,就会出错。所以必须要分为两个范围。 [preorderBegin, preorderEnd] [inorderBegin, inorderEnd] 这两个范围。 第二个错误,错的更深,其实到最后都没有彻底意识到,一定要当心,做此类题时。