1vector<int> preorderTraversal(TreeNode*root) {2vector<int>rVec;3if(!root)4returnrVec;56stack<TreeNode *>st;7st.push(root);8while(!st.empty())9{10TreeNode *tree =st.top();11rVec.push_back(tree->val);12st.pop();13if(tree->right)14st.push(tree->right);15if(tree->left)1...
和上面要求一样,只是要返回以中序方式序列的元素,这次用递归实现: 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:12vector<int> pre...
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...
TreeNode* root= new TreeNode(preorder[preorder_left]); int left_size = inorder_root - inorder_left; //获取左子树的大小 //preorder_left+1---preorder_left+left_size是preorder的左子树位置,对应inordered左子树位置为inorder_left---inorder_root-1。preorder边界是来确定左子树的左右子树边界,...
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。
Return the root of the constructed binary tree. 3.1.2 Indices as parameters: Start by defining a helper function build_tree_helper that takes 4 parameters inorder and postorder's start and end respectively . Identify the root of the binary tree using the last element of the postorder list. ...
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...
Nodes that have no children are referred to asleaf nodes. Nodes that have one or two children are referred to asinternal nodes. Using these new definitions, the leaf nodes in binary tree (a) are nodes 6 and 8; the internal nodes are nodes 1, 2, 3, 4, 5, and 7. ...
Nodes that have no children are referred to asleaf nodes. Nodes that have one or two children are referred to asinternal nodes. Using these new definitions, the leaf nodes in binary tree (a) are nodes 6 and 8; the internal nodes are nodes 1, 2, 3, 4, 5, and 7. ...
classSolution:definorderTraversal(self,root:TreeNode)->List[int]:ifroot==None:return[]stack=[[root,True]]result=[]whilestack:# print('\n---')# print(result)# for s in stack:# print(s[0].val,s[1])cmp1=stack[-1][0].left!=Nonecmp2=stack[-1][0].right!=Noneifnotcmp1andnot...