【LeetCode】(94)Binary Tree Inorder Traversal(Easy)

题目

Binary Tree Inorder Traversal

Total Accepted: 78083 Total
Submissions: 214829My Submissions

Question
Solution

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]

.

解析

没啥说的，中序遍历

/**
* 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:
void inorderTraversalSub_94(TreeNode* root,vector<int>& ret)
{
if (root==NULL)
{
return;
}
inorderTraversalSub_94(root->left,ret);
ret.push_back(root->val);
inorderTraversalSub_94(root->right,ret);
}
vector<int> inorderTraversal(TreeNode* root) {
vector<int> ret;
inorderTraversalSub_94(root,ret);
return ret;
}
};

感觉很简单的代码，但是看了一下大神的，貌似很复杂的样子，难道大神的代码效率更高？但是我用的时间也是接近于0的。

/**
* 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)
{
vector<int> result;
const TreeNode *p = root;
stack<const TreeNode *> s;
while (!s.empty() || p != nullptr)
{
if (p != nullptr)
{
s.push(p);
p = p->left;
} else {
p = s.top();
s.pop();
result.push_back(p->val);
p = p->right;
}
}
return result;
}
};