【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;  }};