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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