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? Hide Tags Tree Stack   一题后续遍历树的问题,很基础,统计哪里的…

Binary Tree Postorder Traversal Given a binary tree, return the postorder traversal of its nodes' values. For example:Given binary tree {1,#,2,3},   1    \     2    /   3return [3,2,1]. Note: Recursive solution is trivial, could you do it iteratively…

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? 经典题目,求二叉树的后序遍历的非递归方法,跟前序,中序,层序一样都需要用到栈,后续的…

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? vector<int> postorderTraversal(TreeNode *r…

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? 后序遍历:左孩子->右孩子->根节点 后序遍历最关键的是利用一个指针保存前一个访问过的信…

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? 中文:二叉树的兴许遍历(左-右-根).能用非递归吗? 递归: public clas…

[题目] 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? [题意] 非递归实现兴许遍历 [思路] 维护两个栈,一个栈用来存储标记,标…

题意: 用迭代法输出一棵二叉树的后序遍历结果. 思路: (1)用两个栈,一个存指针,一个存标记,表示该指针当前已经访问过哪些孩子了. /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solut…

# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def postorderTraversal(self, root): """ :type root: TreeNode :rtype: List[…

先给出递归版本的实现方法,有时间再弄个循环版的.代码如下: /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<int> postorderTraversal…