https://leetcode.com/problems/binary-tree-postorder-traversal/ Given a binary tree, return thepostordertraversal of its nodes' values. Example: [1,null,2,3] [3,2,1] Follow up: Recursive solution is trivial, coul
voidpostOrderTraversalIterative(BinaryTree *root){ if(!root)return; stack<BinaryTree*>s; s.push(root); BinaryTree *prev=NULL; while(!s.empty()){ BinaryTree *curr=s.top(); // we are traversing down the tree if(!prev||prev->left==curr||prev->right==curr){ if(curr->left){ s....
Every time we recursively call one of our traversal methods (e.g., in_order(root.left)) we add a frame to the call stack. When calculations within the frame have been completed, that frame is popped off the call stack and the next line of code is run on the previous frame in the ...
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...
usingnamespacestd; // Data structure to store a binary tree node structNode { intdata; Node*left=nullptr,*right=nullptr; Node(){} Node(intdata):data(data){} }; // Function to print the inorder traversal on a given binary tree ...
vector<int> postorderTraversal(TreeNode*root) {if(!root)return{}; vector<int>res; stack<TreeNode*>s1, s2; s1.push(root);while(!s1.empty()) { TreeNode*t =s1.top(); s1.pop(); s2.push(t);if(t->left) s1.push(t->left);if(t->right) s1.push(t->right); ...
For each test case, print in one line the first number of the postorder traversal sequence of the corresponding binary tree. Sample Input: 7 1 2 3 4 5 6 7 2 3 1 5 4 7 6 Sample Output: 3 坑点: 1. 唯一要注意的是递归太深,会导致堆栈溢出,所以要控制递归返回条件 ...