Binary Tree Inorder Traversal -- LeetCode 94

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].

Solution 1:

/**
 * Definition for a binary tree node.
 * 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)
    {
        if(root != NULL)
        {
            inorderTraversal(root->left);
            v.push_back(root->val);
            inorderTraversal(root->right);
        }
        return v;
    }
private:
    vector<int> v;
};

Solution 2:

时间: 2024-11-01 14:10:49

Binary Tree Inorder Traversal -- LeetCode 94的相关文章

Binary Tree Inorder Traversal leetcode java

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? 题解: 中序遍历:递归左 处理当前 递归右. 画图的话就是,之前离散老师教的,从ro

Binary Tree Inorder Traversal ——LeetCode

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? 题目大意:中序遍历一个二叉树,递归的方案太low,用迭代的方式来写? 解题思路:不用递归,那就

Binary Tree Inorder Traversal [leetcode] 非递归的三种解法

第一种方法是Morris Traversal 是O(n)时间复杂度,且不需要额外空间的方法.缺点是需要修改树. 通过将叶子节点的right指向其中序后继. 代码如下 vector<int> inorderTraversal(TreeNode *root) { vector<int> res; TreeNode * cur = root; TreeNode * pre = NULL; while (cur) { if (cur->left == NULL) { res.push

94. Binary Tree Inorder Traversal 做题报告

题目链接: 94. Binary Tree Inorder Traversal 题目大意: 二叉树的中序遍历 做题报告: (1)该题涉及的算法,数据结构以及相关知识点 递归 (2)自己的解答思路+代码+分析时间和空间复杂度 递归思路 /** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) {

LeetCode Binary Tree Inorder Traversal

LeetCode解题之Binary Tree Inorder Traversal 原题 不用递归来实现树的中序遍历. 注意点: 无 例子: 输入: {1,#,2,3} 1 2 / 3 输出: [1,3,2] 解题思路 通过栈来实现,从根节点开始,不断寻找左节点,并把这些节点依次压入栈内,只有在该节点没有左节点或者它的左子树都已经遍历完成后,它才会从栈内弹出,这时候访问该节点,并它的右节点当做新的根节点一样不断遍历. AC源码 # Definition for a binary tree node

leetcode -day29 Binary Tree Inorder Traversal &amp; Restore IP Addresses

1.  Binary Tree Inorder Traversal 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? 分析:求二叉树的中序

[LeetCode][JavaScript]Binary Tree Inorder Traversal

Binary Tree Inorder Traversal 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? https://leetcode.

leetcode - Binary Tree Preorder Traversal &amp;&amp; Binary Tree Inorder Traversal &amp;&amp; Binary Tree Postorder Traversal

简单来说,就是二叉树的前序.中序.后序遍历,包括了递归和非递归的方法 前序遍历(注释中的为递归版本): 1 #include <vector> 2 #include <stack> 3 #include <stddef.h> 4 #include <iostream> 5 6 using namespace std; 7 8 struct TreeNode 9 { 10 int val; 11 TreeNode *left; 12 TreeNode *rig

LeetCode: Binary Tree Inorder Traversal [094]

[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? >