[LeetCode]144. Binary Tree Preorder Traversal--二叉树前序遍历

> [144. Binary Tree Preorder Traversal](https://leetcode.com/problems/binary-tree-preorder-traversal/description/)

Given a binary tree, return the preorder traversal of its nodes’ values.

For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

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分析：二叉树的基础知识
实现三种遍历方法：
1.递归，时间复杂度：O(n)，空间复杂度：O(n)
2.迭代，时间复杂度：O(n)，空间复杂度：O(n)
3.线索二叉树，时间复杂度：O(n)，空间复杂度：O(1)

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// 解法一：递归/** * 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> preorderTraversal(TreeNode* root) {        vector<int> result;        preTraversal(root, result);        return result;    }    void preTraversal(TreeNode* root, vector<int>& result){        if(!root)            return;        result.push_back(root->val);        preTraversal(root->left, result);        preTraversal(root->right, result);    }};

// 解法二：迭代/** * 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> preorderTraversal(TreeNode* root) {        vector<int> result;        if(!root)            return result;        stack<TreeNode*> s;        TreeNode* temp = root;        while(temp || !s.empty()){            if(temp){                result.push_back(temp->val);                if(temp->right)                    s.push(temp->right);                temp = temp->left;            }else{                temp = s.top();                s.pop();            }        }        return result;    }};

// 解法三：线索二叉树/** * 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> preorderTraversal(TreeNode* root) {        vector<int> v;        if(!root) return v;            TreeNode* temp = root, *prev;        while(temp){            if(!temp->left){                v.push_back(temp->val);                temp = temp->right;            }else{                prev = temp->left;                while(prev->right&&(prev->right != temp))                    prev = prev->right;                if(!prev->right){                    v.push_back(temp->val);                    prev->right = temp;                    temp = temp->left;                }else{                    prev->right = NULL;                    temp = temp->right;                }            }        }        return v;    }};