Suppose that all the keys in a binary tree are distinct positive integers. Given the preorder and inorder traversal sequences, you are supposed to output the first number of the postorder traversal sequence of
pNode=stack.back(); stack.push_back(pNode->right); isRight.push_back(true); } }returnans; } 仅用一个栈的方法https://oj.leetcode.com/discuss/14118/post-order-traversal-using-two-satcks vector<int> postorderTraversal(TreeNode *root) { stack<TreeNode *>st; vector<int>vRet; TreeNode...
Space: O(n). The stack would store all the tree elements in worst case. The set would store all the nodes at the end. 2.2 Iterative solution A better way is based on the fact that post-order traversal is the reverse order of pre-order traversal with visiting the right subtree before ...
def postorderTraversal(self, root): if not root: return [] res = [] stack = [[root,0]] while stack: node = stack[-1] stack.pop() if node[1]== 0 : current = node[0] stack.append([current,1]) if current.right: stack.append([current.right,0]) if current.left: stack.appen...
Implement common Depth-First Traversal (DFS) patterns with recursion and learn about the call stack in this visual guide.def in_order(root=None): if root: in_order(root.left) print(root.val) in_order(root.right) def pre_order(root=None): if root: print(root.val) pre_...
Postorder Traversal: Sample Solution: Java Code: classNode{intkey;Nodeleft,right;publicNode(intitem){// Constructor to create a new Node with the given itemkey=item;left=right=null;}}classBinaryTree{// Root of Binary TreeNoderoot;BinaryTree(){// Constructor to create an empty binary treero...
* Returns an unmodifiable iterable over the nodes in a tree structure, using post-order * traversal. That is, each node's subtrees are traversed before the node itself is returned. * * No guarantees are made about the behavior of the traversal when nodes change while iteration * is...
vector<int> postorderTraversal(TreeNode *root) { stack<TempNode *> s; vector<int> path; TreeNode *p = root; TempNode *temp; while(p !=NULL|| !s.empty()) { while(p !=NULL)//沿左子树一直往下搜索,直至出现没有左子树的结点
vector<int> postorderTraversal(TreeNode*root) { vector<int>res; stack<TreeNode*>_stack; TreeNode*cur=root; TreeNode*pre=NULL;if(cur!=NULL) _stack.push(cur);while(!_stack.empty()) { cur=_stack.top();if((cur->left==NULL&&cur->right==NULL)||(pre&&(cur->left==pre||cur->right...
Riferimenti: https://en.wikipedia.org/wiki/Tree_traversal Esercizio: Esegui l'attraversamento iterativo del post-ordine utilizzando una sola stack. Vota questo post Voto medio 4.61/5. Conteggio voti: 261 Grazie per aver letto. Si prega di utilizzare il nostro compilatore in linea per pub...