代码如下: 1#Definition for a binary tree node.2#class TreeNode(object):3#def __init__(self, x):4#self.val = x5#self.left = None6#self.right = None78classSolution(object):9defpostorderTraversal(self, root):10"""11:type root: TreeNode12:rtype: List[int]13"""14res =[]15tmp ...
代码如下: 1/**2* Definition for binary tree3* struct TreeNode {4* int val;5* TreeNode *left;6* TreeNode *right;7* TreeNode(int x) : val(x), left(NULL), right(NULL) {}8* };9*/10classSolution {11public:12vector<int> postorderTraversal(TreeNode *root) {13//IMPORTANT: Please...
vector<int>postorderTraversal(TreeNode*root) { vector<int>res; if(root==NULL){ returnres; } stack<TreeNode*>stackNode; stackNode.push(root); TreeNode*preNode=NULL; while(stackNode.empty()==false){ TreeNode*curNode=stackNode.top(); boolisLeafNode=curNode->left==NULL&&curNode->right=...
*/voidtraversal(structTreeNode*root,int*countPointer,int*res){if(!root)return;traversal(root->left,countPointer,res);traversal(root->right,countPointer,res);res[(*countPointer)++]=root->val;}int*postorderTraversal(structTreeNode*root,int*returnSize){int*res=malloc(sizeof(int)*110);intcount=...
Given a binary tree, return the postorder traversal of its nodes’ values. For example: Given binary tree{1,#,2,3}, 1 \ 2 / 3 1. 2. 3. 4. 5. return[3,2,1]. Note: Recursive solution is trivial, could you do it iteratively?
此题来自leetcode https://oj.leetcode.com/problems/binary-tree-postorder-traversal/ Given a binary tree, return the postorder traversal of its nodes' values. For example: Given binary tree{1,#,2,3}, 代码语言:javascript 复制 1\2/3
Binary Tree Postorder Traversal 思路: 递归法解决,迭代法过于麻烦,按下不表。 代码: publicList<Integer>postorderTraversal(TreeNoderoot){List<Integer>res=newArrayList<>();inorder(res,root);returnres;}privatevoidinorder(List<Integer>res,TreeNodenode){if(node==null){return;}inorder(res,node.left)...
traversal(root) return res 结果: 后序遍历 题目:Binary Tree Postorder Traversal - LeetCode 代码: class TreeNode: def __init__(self, val=0, left=None, right=None): self.val = val self.left = left self.right = right class Solution: ...
1. Problem Descriptions:Given two integer arrays inorderandpostorderwhereinorderis the inorder traversal of a binary tree andpostorderis the postorder traversal of the same tree, construct and retu…
=null){stack.push(cur);cur=cur.left;}cur=stack.pop();list.add(cur.val);cur=cur.right;}returnlist;}//后序遍历,非递归publicstaticList<Integer>postOrder(TreeNoderoot){Stack<TreeNode>stack=newStack<>();List<Integer>list=newArrayList<>();TreeNodecur=root;TreeNodep=null;//用来记录上一节点...