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, could you do it iteratively? 回到顶部 Intuition Method 1. Using one stack to s...
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....
If the node has a right child, push the right child onto the stack. If the node has a left child, push the left child onto the stack. Repeat steps 3-6 until the stack is empty. function preorderTraversal(root) { if (!root) return []; const stack = [root]; ...
This space is primarily used for the recursive call stack during the construction of the tree. Additionally, the inorder_index dictionary requires O(n) space, as it stores the index of each value in the inorder traversal list. Overall, the space usage is proportional to the number of nodes...
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 ...
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 思路: 这个题目要求输出后序遍历第一个元素,所以只需要进行分割和输出即可。为了尽快推出递归,增加...