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 =newHash
* 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: voiddfs(vector<int>&preorder,TreeNode*root,intleft,intright){ if(left>right) ...
} TreeNode* bstFromPreorder(vector<int>&preorder) {returndfs(0, preorder.size() -1, preorder); } }; 2. O(N) classSolution {public:inti =0; TreeNode* bstFromPreorder(vector<int>& A,intbound =INT_MAX) {if(i == A.size() || A[i] > bound)returnnullptr; TreeNode* root =...
(Recall that a binary search tree is a binary tree where for everynode, any descendant ofnode.lefthas a value<node.val, and any descendant ofnode.righthas a value>node.val. Also recall that a preorder traversal displays the value of thenodefirst, then traversesnode.left, then traversesn...
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...
Memory Usage: 35.8 MB, less than 99.68% of Java online submissions for Construct Binary Search Tree from Preorder Traversal. 网友最佳解法1,最优解法是以插入方式去拼出整个树 public TreeNode bstFromPreorder(int[] preorder) { TreeNode root = new TreeNode(preorder[0]); for (int i = 1; ...
* TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{public TreeNodebuildTree(int[]preorder,int[]inorder){if(preorder==null||preorder.length==0)returnnull;if(inorder==null||inorder.length==0)returnnull;if(preorder.length!=inorder.length...
Suppose we have to create a binary search tree that matches the given preorder traversal. So if the pre-order traversal is like [8,5,1,7,10,12], then the output will be [8,5,10,1,7,null,12], so the tree will be − To solve this, we will follow these steps − root :=...
preorder traversal of a tree * @param inorder: A list of integers that inorder traversal of a tree * @return: Root of a tree */ public void main(String[] args){ int[] A= new int[] {2,1,3} ; int[] B = new int[] {1,2,3}; TreeNode root = buildTree(A , B) ; ...
Construct Binary Tree from Preorder and Inorder Traversal Given preorder and inorder traversal of a tree, construct the binary tree. 这题跟上面那题类似,通过前序遍历和中序遍历的结果构造二叉树,我们只需要知道前序遍历的第一个值就是根节点,那么仍然可以采用上面提到的方式处理: ...