中序遍历(Inorder Traversal)是二叉树遍历的一种方法,其遍历顺序为:先遍历左子树,然后访问根节点,最后遍历右子树。这种遍历方式被广泛应用于二叉树的各类算法中,特别是当需要按照某种特定顺序处理二叉树节点时,中序遍历往往能发挥重要作用。通过中序遍历,可以确保二叉树的每个节点都被且仅...
class Solution{public:vector<int>inorderTraversal(TreeNode*root){vector<int>r;if(root==NULL){returnr;}stack<TreeNode*>st;TreeNode*p=root;while(p||!st.empty()){while(p){st.push(p);p=p->left;}p=st.top();st.pop();r.push_back(p->val);p=p->right;}returnr;}}; ...
Preorder traversal starts printing from the root node and then goes into the left and right subtrees, respectively, while postorder traversal visits the root node in the end. #include<iostream>#include<vector>using std::cout;using std::endl;using std::string;using std::vector;structTreeNode{...
intmain(){//Binary tree traversalBinTree tree; Node h = {8,NULL,NULL}; Node g = {7,NULL, &h}; Node f = {6,NULL,NULL}; Node e = {5, &f, &g}; Node d = {4,NULL, &e}; Node c = {3,NULL,NULL}; Node b = {2, &d,NULL}; Node a = {1, &b, &c}; tree.Set...
So let's traverse the below tree usingreverse inordertraversal. For the above tree, the root is:7 Traverse the right subtree first(subtree rooted by 9) Now to traverse the subtree rooted by 9, it again follows the same set of rules ...
*/voidtraversal(structTreeNode*root,int*countPointer,int*res){if(!root)return;traversal(root->left,countPointer,res);res[(*countPointer)++]=root->val;traversal(root->right,countPointer,res);}int*inorderTraversal(structTreeNode*root,int*returnSize){int*res=malloc(sizeof(int)*110);intcount=...
Binary search tree traversal in order, postorder, and preorder traversal. Top of the tree, the height of the tree inorder-traversalpreorder-traversalpostorder-traversaltop-view-binary-treeheight-of-tree UpdatedSep 2, 2020 Python jaydattpatel/Binary-Tree ...
and finally nodes on the right subtree. You start traversal from root then go to the left node, then again go to the left node until you reach a leaf node. At that point in time, you print the value of the node or mark it visited and move to the right subtree. Continuing the same...
vector<int> inorderTraversal(TreeNode *root) { // Start typing your C/C++ solution below // DO NOT write int main() function vector<int> * vec = new vector<int>(); stack<TreeNode*> my_stack; while(1) { if( root) { vec->push_back(root->val); ...
用递归比较方便。public class Solution { public TreeNode buildTree(int[] preorder, int[] inorder) { if (preorder.length != inorder.length) return null; return