Given a binary search tree, print the elements in-order iteratively without using recursion.Note:Before you attempt this problem, you might want to try coding a pre-order traversal iterative solution first, because it is easier. On the other hand, coding a post-order iterative version is a ...
vector<int> inorderTraversal(TreeNode* root) { vector<int> ans; stack<pair<TreeNode*,int>> s; /*will have to maintain if left subtree has been checked for this node or not 0 for not checked 1 for checked*/ if(root){ ...
classSolution(object):definorderTraversal(self, root):#递归""":type root: TreeNode :rtype: List[int]"""ifnotroot:return[]returnself.inorderTraversal(root.left) + [root.val] + self.inorderTraversal(root.right) Java解法 classSolution {publicList < Integer >inorderTraversal(TreeNode root) ...
还可以通过iterative实现 Iterative Preorder Traversal - GeeksforGeekswww.geeksforgeeks.org/iterative-preorder-traversal/ Iterative Postorder Traversal | Set 2 (Using One Stack) - GeeksforGeekswww.geeksforgeeks.org/iterative-postorder-traversal-using-stack/ 这个postorder有点tricky,需要考虑比如最左...
preorder(copy tree), inorder (non-descending order for BST), postorder traversal (delete tree) 的recursive方法还有iterative的方法 http://www.geeksforgeeks.org/618注意比较三种code,pre order 和inorder仅仅是res赋值的位置不一样,postorder pre order不停地遍历左子树,并且每次遍历都给res赋值,都push到st...
2014-1-9 update iterative的解法,我使用一个标志flag,判断是否遍历过的节点: vector<int> inorderTraversal(TreeNode *root) { vector<int> rs; if (!root) return rs; stack<TreeNode *> stk; stk.push(root); bool flag = false; while (!stk.empty()) ...
public List<Integer> inorderTraversalIterative(TreeNode root) { List<Integer> result = new ArrayList<Integer>(); if(root==null){ return result; } Stack<TreeNode> stack = new Stack<TreeNode>(); TreeNode current = root; while(!stack.isEmpty() || current!=null){ if(current != null)...
The final traversal order is 1, 3, 9, 2, 17, 25, 90, 57, 13, 11. Pseudocode Iterative Method: Insert(T, z) y:=nil x:=T.rootwhilex!=nil loop y:=xifz.key<x.key then x:=x.leftelsex:=x.right endifend loop z.parent:=y--x reaches nil, so y is a leaf and becomes par...
We have to find the post order traversal of this tree using the iterative approach. So if the tree is like − Then the output will be: [9,15,7,10,-10] To solve this, we will follow these steps − if root is null, then return empty array create an array ret stack := define...
(1.helper-null,left,add,right; 2.inorderTraversal-rst,helper,return;) View Code Iterative解法:<Stack>(4-2-6-1-3-5-null, 当cur!=null时, cur=cur.left遍历左子树, push; 当cur为null时跳出内while循环, 此时cur对于1的左子结点即null, stack里为1,2,4; pop1,cur=1; 接着cur=cur.right;...