}voidhelp(vector<int> &vec, TreeNode*root){if(root == nullptr)return; help(vec, root->left); help(vec, root->right); vec.push_back(root->val); } }; //迭代classSolution {public: vector<int> postorderTraversal(TreeNode*root) {if(root == nullptr)return{}; vector<int>res; stack...
void PreOrderPrintLeaves ( BinTree BT ) { if ( BT ) { if ( !BT->Left && !BT->Right ) printf("%c", BT->Data); PreOrderPrintLeaves( BT->Left ); PreOrderPrintLeaves( BT->Right ); } } 复制代码 2.后序遍历求二叉树的高度 int PostOrderGetHeight ( BinTree BT) { int HL, HR,...
1、递归 1structTreeNode{2TreeNode *left;3TreeNode *right;4intval;5TreeNode(intx):val(x),left(NULL),right(NULL){}6}7vector<int> postorder(treeNode*root){8vector<int>res;9postordertraversal(root,res);10returnres;11}1213voidpostorderTraversal(TreeNode *root,vector<int> &res){14if(root...
publicList<Integer>postorderTraversal(TreeNoderoot){List<Integer>list=newArrayList<>();Stack<TreeNode>stack=newStack<>();TreeNodecur=root;TreeNodelast=null;while(cur!=null||!stack.isEmpty()){if(cur!=null){stack.push(cur);cur=cur.left;}else{TreeNodetemp=stack.peek();//是否变到右子树if...
145. Binary Tree Postorder Traversal 二叉树的后序遍历 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。
Leetcode 之Binary Tree Postorder Traversal(44) 后序遍历,比先序和中序都要复杂。访问一个结点前,需要先判断其右孩子是否被访问过。如果是,则可以访问该结点;否则,需要先处理右子树。 vector<int> postorderTraversal(TreeNode *root) { vector<int>result;...
Postorder traversal Essentially, all three traversals work in roughly the same manner. They start at the root and visit that node and its children. The difference among these three traversal methods is the order with which they visit the node itself versus visiting its children. To help clarify...
Postorder traversal Essentially, all three traversals work in roughly the same manner. They start at the root and visit that node and its children. The difference among these three traversal methods is the order with which they visit the node itself versus visiting its children. To help clarify...
title: binary-tree-postorder-traversal 描述 Given a binary tree, return thepostordertraversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 return[3,2,1]. Note:Recursive solution is trivial, could you do it iteratively?
binary-tree-postorder-traversal publicvector<int>postorderTraversal(TreeNode*root){vector<int>result;stack<TreeNode*>st;TreeNode*p=root,*q=NULL;do{while(p)st.push(p),p=p->left;q=NULL;while(!st.empty()){p=st.top();st.pop();if(p->right==q)//右子树为空或者已经访问{result.push_...