}/*** LNR:中序遍历(Inorder Traversal) * 访问根结点的操作发生在遍历其左右子树之中(间)。*/publicvoidinOrderTraversal() {//Lif(this.leftSubNode !=null) {this.leftSubNode.inOrderTraversal(); }//NSystem.out.print(this);//Rif(this.rightSub
* int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * };*/classSolution {public: vector<int> inorderTraversal(TreeNode*root) { vector<int>ret;if( !root )returnret; stack<TreeNode*>sta; sta.push(root);while( !sta.empty...
inorder(root -> right,vec); } } vector<int> inorderTraversal(TreeNode* root) { vector<int> vec; if(root == NULL)return vec; inorder(root,vec); return vec; } }; 递归 class Solution { public: void inorder(TreeNode* root, vector<int> & vec) { if(root -> left != NULL) {...
2.2 last.right 不为 null,说明之前已经访问过,第二次来到这里,表明当前子树遍历完成,保存 cur 的值,更新 cur = cur.right public List<Integer> inorderTraversal3(TreeNode root) { List<Integer> ans = new ArrayList<>(); TreeNode cur = root; while (cur != null) { //情况 1 if (cur.left ...
Given a binary tree, return thepreordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 1. 2. 3. 4. 5. return[1,2,3]. 解法一: 递归方法: 如果root不为空, 先访问根,递归左子树,递归右子树。
* TreeNode(int x) { val = x; } * } */classSolution{public List<Integer>inorderTraversal(TreeNode root){List<Integer>result=newLinkedList<>();TreeNode current=root;TreeNode prev=null;while(current!=null){// left firstif(current.left==null){result.add(current.val);current=current.right...
Given binary tree [3,9,20,null,null,15,7], 3 / \ 9 20 / \ 15 7 return its bottom-up level order traversal as: [[15,7],[9,20],[3]] 二叉树的层次遍历,不过需要保持每一层的数据,可以通过vector配合队列实现,实现: classSolution{public:vector<vector<int>>levelOrderBottom(TreeNode*root...
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. 方法:在inorder中寻找postorder的最后一个,然后左右递归,注意处理好各个low,high边界 1classSolution2{3public:4TreeNode *buildTree(vector<int> &inorder, vect...
The pre-order and post-order traversal of a Binary Tree is A 、B 、C and C 、B 、A, respectively. The number of trees that satisfy the conditions is . A.1 B.2 C.3 D.4 暂无答案
levelOrderTraversal(root); return0; } DownloadRun Code Output: 15 10 20 8 12 16 25 The time complexity of the above solution isO(n2), wherenis the total number of nodes in the binary tree. The auxiliary space required by the program isO(h)for the call stack, wherehis the height of...