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->...
*/// 由于主函数的形式已经符合分治法需要的形式(具有合适的返回值),直接使用主函数做为递归函数vector<int>preorderTraversal(TreeNode * root){//递归三要素之定义// write your code herevector<int> result;if(root == nullptr) {returnresult;// 递归三要素之出口}// 获取左子树的遍历结果vector<int> ...
classSolution {public: vector<int> preorderTraversal(TreeNode *root) { vector<int>res; stack<TreeNode *>s;if(root == NULL){//空树returnres; } s.push(root);//放树根while(!s.empty()){ TreeNode*cc =(); s.pop(); res.push_back(cc->val);//访问根if(cc->right !=NULL){ s.p...
12 vector<int> preorderTraversal(TreeNode *root) { 13 vector<int> v; 14 if(root==NULL) 15 return v; 16 stack<TreeNode *> tree; 17 tree.push(root); 18 while(!tree.empty()){ 19 TreeNode *p=tree.top(); 20 v.push_back(p->val); 21 tree.pop(); 22 if(p->right!=NULL) 2...
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。 public List<Integer> preorderTraversal(TreeNode root) { Li...
Morris traversal: My code: /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{publicList<Integer>preorderTraversal(TreeNoderoot){List<Integer>ret=newArrayList<...
public:vector<int>preorderTraversal(TreeNode*root){vector<int>result;stack<TreeNode*>st;if(root==NULL)returnresult;st.push(root);while(!st.empty()){TreeNode*top=st.top();st.pop();result.push_back(top->val);if(top->right)st.push(top->right);if(top->left)st.push(top->left);}...
Binary tree traversal: Preorder, Inorder, and Postorder In order to illustrate few of the binary tree traversals, let us consider the below binary tree: Preorder traversal: To traverse a binary tree in Preorder, following operations are carried-out (i) Visit the root, (ii) Traverse the le...
Pre-order traversal in a tree Previous Quiz Next In this traversal method, the root node is visited first, then the left subtree and finally the right subtree. We start from A, and following pre-order traversal, we first visit A itself and then move to its left subtree B. B is also...
The preorder traversal sequence of a binary tree is abdgcefh, the inorder traversal sequence of a binary tree is dgbaechf, the postorder traversal sequence of a binary tree is ___. A. bdgcefha B. gdbecfha C. bdgaechf D. gdbehfca ...