TreeNode *pNodeA6 = CreateBinaryTreeNode(4); TreeNode *pNodeA7 = CreateBinaryTreeNode(7); ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3); ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5); ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7); PrintTree(pNodeA1); vector<int> ans =postorderTraversa...
= left; * this.right = right; * } * } */ class Solution { List<Integer> result = new LinkedList<>(); public List<Integer> postorderTraversal(TreeNode root) { traversal(root); return result; } private void traversal(TreeNode root) { if (root == null) { return; } traversal(root...
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....
publicList<Integer>postorderTraversal(TreeNoderoot){List<Integer>list=newArrayList<>();if(root==null){returnlist;}Stack<TreeNode>stack=newStack<>();stack.push(root);stack.push(root);while(!stack.isEmpty()){TreeNodecur=stack.pop();if(cur==null){continue;}if(!stack.isEmpty()&&cur==stac...
* TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: //非递归方法实现后序遍历 vector<int> postorderTraversal(TreeNode* root) { vector<int> res; if(root==NULL) return res; stack<TreeNode *> vis; ...
给定一个二叉树,返回它的后序遍历。 示例: 输入:[1,null,2,3] 1 \ 2 / 3输出:[3,2,1] 1. 2. 3. 4. 5. 6. 7. 8. 进阶:递归算法很简单,你可以通过迭代算法完成吗? DFS 今天这道题比较简单,直接DFS就可以了。 Code def postorderTraversal(self, root: TreeNode) -> List[int]: ...
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* public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */classSolution{public List<Integer>postorderTraversal(TreeNode root){Deque<Integer>result=newLinkedList<>();if(root==null){returnnewArrayList(result);}Stack<TreeNode>tracker...
public void dfs(TreeNode node){ if(node == null) return; dfs(node.left); dfs(node.right); res.add(node.val); } } class Solution { public static List<Integer> postorderTraversal(TreeNode root) { Deque<TreeNode> stack = new LinkedList<>(); ...
Binary tree traversal: Preorder, Inorder, and Postorder In order to illustrate few of the binary tree traversals, let us consider the below binary tree: Preorder traversal: To traverse a binary tree in Preorder, following operations are carried-out (i) Visit the root, (ii) Traverse the le...