Binary Tree Preorder Traversal--二叉树的先序遍历

原题：

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

=>给出一个二叉树，返回先序遍历的所有的节点值

For example:

例如：
Given binary tree {1,#,2,3},

给出下面的二叉树

   1    \     2    /   3

return [1,2,3].

返回[1,2,3]

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

=》注意：递归的算法是很普通，能不能不要递归呢？

/** * Definition for binary tree * 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) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.            }};

 

晓东解析：

这个题目用递归来实现的确很简单，就是先遍历左，再遍历右。那非递归的算法就是先一直左到底，并把这些所有的左都压到一个栈里面，然后到最后，再一个个pop出来，每pop一个，对它的右子树再进行一次类似的遍历就可以了。

 

代码实现：

1、递归实现

/*** Definition for binary tree* 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) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        vector<int> result;        vector<int> left;        vector<int> right;                if(root == NULL) return result;        result.push_back(root->val);        left = preorderTraversal(root->left);        right = preorderTraversal(root->right);                if(left.size() != 0)            result.insert(result.end(), left.begin(), left.end());        if(right.size() != 0)            result.insert(result.end(), right.begin(), right.end());                return result;            }};

运行结果：

67 / 67test cases passed.

Status:

Accepted

Runtime: 20 ms

 

2、非递归实现：

/** * Definition for binary tree * 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) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.        stack<TreeNode*> TreeStack;        vector<int> result;                if(root == NULL) return result;                while(root || !TreeStack.empty()){            while(root){                TreeStack.push(root);                result.push_back(root->val);                root = root->left;            }            root = TreeStack.top();            TreeStack.pop();            root = root->right;        }            }};

运行结果：

67 / 67test cases passed.

Status:

Accepted

Runtime: 12 ms

 

 

希望大家有更好的算法能够提出来，不甚感谢。

 

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