map<int, int> mapIndex; void mapToIndex(int inorder[], int n) { for (int i = 0; i < n; i++) { mapIndex.insert(map<int, int>::value_type(inorder[i], i); } } Node* buildInorderPreorder(int in[], int pre[], int n,
再进行递归就可以解决问题了。 1/**2* Definition for binary tree3* 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:12TreeNode *buildTree(vector<int> &inorder, vector<...
0,preorder.length,inorder,0,inorder.length);}privateTreeNodebuildTreeHelper(int[]preorder,intp_start,intp_end,int[]inorder,inti_start,inti_end){// preorder 为空,直接返回 nullif(p_start==p_end){returnnull;}introot_val=preorder[p_start];TreeNoderoot=newTreeNode(root_val);//在中序遍...
3. Main Function (buildTree): If the inorder list is empty, we return None as there are no elements to build the tree from. We call the helper function build_tree_helper with initial boundaries 0 (leftmost) and len(inorder) - 1 (rightmost) to build the entire tree. 4. Return Value...
preorder = [3,9,20,15,7] inorder = [9,3,15,20,7] 1. 2. Return the following binary tree: 3 / \ 9 20 / \ 15 7 1. 2. 3. 4. 5. 题目大意 使用先序遍历和中序遍历序列重构二叉树。 解题方法 递归 解决本题需要牢牢掌握先序遍历和中序遍历的含义,以及递归。
Similar to pre-order and in-order traversal, accessing the binary tree in the order of left child->right child->root node is called post-order traversal. The first visited node in the post-order traversal is与先序、中序遍历类似,以左子->右子->根节点的顺序来访问二叉树称为后序遍历。后序...
Binary tree (a) has 8 nodes, with node 1 as its root. Node 1's left child is node 2; node 1's right child is node 3. Notice that a node doesn't need to have both a left child and right child. In binary tree (a), node 4, for example, has only a right child. Furthermor...
Binary tree (a) has 8 nodes, with node 1 as its root. Node 1's left child is node 2; node 1's right child is node 3. Notice that a node doesn't need to have both a left child and right child. In binary tree (a), node 4, for example, has only a right child. Furthermor...
树的三种DFS遍历,是指按照根节点(自己)被访问的顺序 Pre-Order: 先访问根节点,再访问其左右子树。对每一个subtree,同样适用。 In-Order: 先访问其...
[preorderBegin]){rootIndex=i;break;}}int leftTreeSize=rootIndex-inorderBegin;root.left=buildTree(preorderBegin+1,preorderBegin+leftTreeSize,inorderBegin,rootIndex-1,preorder,inorder);root.right=buildTree(preorderBegin+leftTreeSize+1,preorderEnd,rootIndex+1,inorderEnd,preorder,inorder);...