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->...
Postorder与Inorder很相似,但是比Inorder复杂的地方是如何判断该节点的左右子树都已经访问过了,按照Inorder的写法左子树还是先被访问,没有问题,但是访问完左子树后不能直接访问当前节点,要判断当前节点的右子树是否已经被访问,如果没有访问则应该继续去访问右子树,最后再访问当前节点 1vector<int> postorderTraversal(T...
1/**2* Definition for a binary tree node.3* 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> preorderTraversal(TreeNode*root) {13vector<int>rVec;14pr...
staticvoidinorder_visit(node Anode) { if(Anode.hasleftchild) { inorder_visit(Anode.leftchild); } Console.Write(Anode.nodevalue); if(Anode.hasrightchild) { inorder_visit(Anode.rightchild); } } //后续遍历-LRD staticvoidpostorder_visit(node Anode) { if(Anode.hasleftchild) { postorder_visi...
inorder_visit(node_a);//中序 Console.WriteLine(); postorder_visit(node_a);//后序 Console.WriteLine(); node node_1 =newnode("1"); node node_2 =newnode("2"); node node_3 =newnode("3"); node node_4 =newnode("4");
顾名思义,在深度优先的情况下,在访问下一个同级树之前,树向下(向深度)遍历,二叉树的PreOrder,InOrder和PostOrder遍历实际上是深度优先遍历。在广度优先上,树的整个宽度在移到下一个级别之前被遍历,因此它也被称为级别顺序遍历。还有其他遍历二叉树的算法,例如蒙特卡洛树搜索,它集中于分析最优的移动,但先序、后序...
然后就是, preorder 与 inorder 的范围不是一致的,必须要分开传入子函数,否则一定会错。 通过postorder 与 inorder 来重构树,也是一个道理。上次做这道题目虽然很轻松地过了,但是一定是瞎猫碰到死耗子,蒙混过关。刚刚说的那些错误点,在这个类型里面同样存在!
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...
The program creates a binary tree for breadth-first traversal.But i'm trying to use Pre-Order, In-Order, Post-Order Traversal and actually i can't do that. The output of the program is not what i expected. I think i should change the preorder, inorder or postorder functions but i ...
Given a binary tree, return the postorder traversal of its nodes' values. Note: Recursive solution is trivial, could you do it iteratively? postorder: left, right, root Recursive version:/** * Definition for a binary tree node. * public class TreeNode { ...