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, coul
Apart from using a Stack, we can use a Set to keep track of all the visited nodes so that we only visit each node once. When deciding whether add current nodes' leftr/right child onto the stack, we first check if it's null or not and then if it's present in the set (which me...
#include <iostream>#include<string>#include<string.h>#include#include<set>#include<list>#include<vector>#include<deque>#include<unordered_set>#include<algorithm>#include<unordered_map>#include<stack>#include<cstdio>usingnamespacestd;intpreArr[50001];intinArr[50001];intn;intpreIndex =1;intflag =...
Implement common Depth-First Traversal (DFS) patterns with recursion and learn about the call stack in this visual guide.def in_order(root=None): if root: in_order(root.left) print(root.val) in_order(root.right) def pre_order(root=None): if root: print(root.val) pre_...
[1:]: insert(Tree, element) return Tree class Solution(object): def postorderTraversal(self, root): if not root: return [] res = [] stack = [[root,0]] while stack: node = stack[-1] stack.pop() if node[1]== 0 : current = node[0] stack.append([current,1]) if current....
1 \ 2 / 3 思路: 借助于一个栈,依次将根节点的右子节点和左子节点压入栈中。如果一个节点为叶子节点,或者前一个出栈的元素为当前栈顶节点的子节点,则出栈。 vector<int> postorderTraversal(TreeNode*root) { vector<int>res; stack<TreeNode*>_stack; ...
vector<int> postorderTraversal(TreeNode *root) { stack<TempNode *> s; vector<int> path; TreeNode *p = root; TempNode *temp; while(p !=NULL|| !s.empty()) { while(p !=NULL)//沿左子树一直往下搜索,直至出现没有左子树的结点
Given a binary tree, return the postorder traversal 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?We know the elements can be printed post-order easily using recursion, as ...
【构建二叉树】02根据中序和后序序列构造二叉树【Construct Binary Tree from Inorder and Postorder Traversal】 我们都知道,已知中序和后序的序列是可以唯一确定一个二叉树的。 初始化时候二叉树为:=== 中序遍历序列,===O=== 后序遍历序列,===O 红色部分是左子树,...
#include<iostream>#include<vector>#include<stack>#include<algorithm>// 引入 reverse函数,对vector进行反转usingnamespacestd;structTreeNode{intval; TreeNode *left; TreeNode *right;TreeNode(intx) :val(x),left(NULL),right(NULL) {} };classSolution{public:vector<int>postorderTraversal(TreeNode* root...