TreeNode *pNodeA6 = CreateBinaryTreeNode(4); TreeNode *pNodeA7 = CreateBinaryTreeNode(7); ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3); ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5); ConnectTreeNodes(pNodeA5, pNode
= 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...
classSolution {public: vector<int> postorderTraversal(TreeNode *root) { vector<int>res;if(root==NULL)returnres; stack<pair<TreeNode*,int>>s;intunUsed; s.push(make_pair(root,1));while(!s.empty()) { root=s.top().first; unUsed=s.top().second; s.pop();if(unUsed){ s.push(make...
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...
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]: ...
Can you solve this real interview question? Binary Tree Preorder Traversal - Given the root of a binary tree, return the preorder traversal of its nodes' values. Example 1: Input: root = [1,null,2,3] Output: [1,2,3] Explanation: [https://assets.l
105 Construct Binary Tree from Preorder and Inorder Traversal 从前序与中序遍历序列构造二叉树,给定一棵树的前序遍历与中序遍历,依据此构造二叉树。注意:你可以假设树中没有重复的元素。例如,给出前序遍历=[3,9,20,15,7]中序遍历=[9,3,15,20,7]返回如下的二叉树:3/\
* 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<>(); ...
postOrderBiTree(T->left); postOrderBiTree(T->right); printf("%c",T->data); } 非递归的方法,以下第一方法便于理解与记忆。第二个方法不便于记忆; 优先记忆此方法: C++Solution1:classSolution{public: vector<int> postorderTraversal(TreeNode*root){if(!root)return{}; ...