* Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */classSolution{public List<Integer>preorderTraversal(TreeNode root){List<Integer>result=newLinkedList<>();TreeNode current=root;TreeNo...
#Definition for a binary tree node.#class TreeNode(object):#def __init__(self, x):#self.val = x#self.left = None#self.right = None"""利用堆栈先进后出的特点"""classSolution(object):defpreorderTraversal(self, root):""":type root: TreeNode :rtype: List[int]"""result=[] slack=...
非递归方法(利用栈) 1/**2* Definition for binary tree3* public class TreeNode {4* int val;5* TreeNode left;6* TreeNode right;7* TreeNode(int x) { val = x; }8* }9*/10importjava.util.*;11publicclassSolution {12publicArrayList<Integer>postorderTraversal(TreeNode root) {13ArrayList<...
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; ...
* Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */// 解法一 递归funcpreorderTraversal(root*TreeNode)[]int{res:=[]int{}ifroot!=nil{res=append(res,root.Val)tmp:=preorderTraversal(root.Left)for_,t:=rangetmp{res=app...
* Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */publicclassSolution{public List<List<Integer>>levelOrder(TreeNode root){if(root==null){returnnewArrayList<List<Integer>>();}List<...
Thus, a tree provides high efficiency for all the common data storage operations: searches, insertions, and deletions. Traversing is not as fast as the other operations, but it must be O(N) to cover all N items, by definition. In all the data structures you’ve seen, it has been O(...
According to the definition, the difference between the height of left sub tree and the right subtree must not be greater than 1. In this case, the difference comes to be 2, which is greater than 1; therefore, the above binary tree is an unbalanced binary search tree.Why do we need a...
二叉树的前序遍历递归实现 /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val...
confused what"{1,#,2,3}"means?> read more on how binary tree is serialized on OJ. 先将结果v存入stack中,最后在从stack倒入res形成倒序,未找到其他好的方法 1/**2* Definition for binary tree3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) :...