话不多说,直接上代码 class TreeNode: def __init__(self, x): self.val = x self.left = None self.right = None class Solution: #前序 def preorder(self,root,ans=[]): if root!=None: ans.append(root.val) if root.left: self.preorder(root.left,ans) if root.right: self.p
#include <stdio.h> #include <stdlib.h> typedef struct tree { int number ; struct tree *left ; struct tree *right ; }TREE; //对树插入节点 void insert_tree(TREE **header , int number) { //创建一颗树 TREE *New = NULL ; New = malloc(sizeof(TREE)); if(NULL ==
Given a binary tree, return the postorder traversal of its nodes' values. Example: Input: [1,null,2,3] 1 \ 2 / 3 Output: [3,2,1] Follow up: Recursive solution is trivial, could you do it iteratively? 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