【题目】
Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3}
,
1 \ 2 / 3
return [1,3,2]
.
Note: Recursive solution is trivial, could you do it iteratively?
confused what "{1,#,2,3}"
means? >
read more on how binary tree is serialized on OJ.
【代码】
/********************************* * 日期:2014-11-17 * 作者:SJF0115 * 题号: Binary Tree Inorder Traversal * 来源:https://oj.leetcode.com/problems/binary-tree-inorder-traversal/ * 结果:AC * 来源:LeetCode * 总结: **********************************/ #include <iostream> #include <malloc.h> #include <vector> #include <stack> using namespace std; struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; class Solution { public: vector<int> inorderTraversal(TreeNode *root) { vector<int> v; if (root == NULL){ return v; } // 根节点入栈 stack<TreeNode*> stack; TreeNode* node = root; // 遍历 while(node != NULL || !stack.empty()){ //遍历左子树 if(node != NULL){ stack.push(node); node = node->left; } else{ //左子树为空,访问右子树 node = stack.top(); stack.pop(); v.push_back(node->val); node = node->right; } } return v; } }; //按先序序列创建二叉树 int CreateBTree(TreeNode* &T){ char data; //按先序次序输入二叉树中结点的值(一个字符),‘#’表示空树 cin>>data; if(data == '#'){ T = NULL; } else{ T = (TreeNode*)malloc(sizeof(TreeNode)); //生成根结点 T->val = data-'0'; //构造左子树 CreateBTree(T->left); //构造右子树 CreateBTree(T->right); } return 0; } int main() { Solution solution; TreeNode* root(0); CreateBTree(root); vector<int> v = solution.inorderTraversal(root); for(int i = 0;i < v.size();i++){ cout<<v[i]<<endl; } }
时间: 2024-11-03 18:27:55