前序遍历: 根->左->右 递归方法 publicvoidpreorder(TreeNode root,List<Integer> res){if(root ==null){return; } res.add(root.val); preorder(root.left,res); preorder(root.right,res);return; }publicList<Integer>preorderTraversal(TreeNode root){ List<Integer> res =newArrayList<Integer>(); preorder(root,res);returnres; ...
*/// 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>ans;if(!root)returnans;stack<TreeNode*>s;s.push(root);while(!s.empty()){root=();s.pop();ans.push_back(root->val);if(root->right)s.push(root->right);if(root->left)s.push(root->left);}returnans;}}...
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...
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...
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);}...
题目: Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 根据前序遍历和中序遍历结果构造二叉树。 思路分析: 分析二叉树前序遍历和中序遍历的结果我们发现: 二叉树前序遍历的第一个节点是根节点。 在中序遍历...
Binary Tree Preorder Traversal 题目描述 分析 二叉树前序遍历,非常简单,不过题目要求用递推法,不可用递归,不经常写的最好再熟悉下,虽然不难,面试中说不定会被问到。递推法只需额外开辟一个栈即可,每遍历完当前节点,按顺序将右孩子(如果有的话)和左孩子(如果有的话)依次入栈,下次遍历则直接从栈顶取就好...
INORDER AND PREORDER TRAVERSAL and DISPLAY OF EXISTING BINARY TREE IN COMPUTER MEMORYP K KumaresanJournal of Harmonized Research