[LeetCode]Binary Tree Preorder Traversal

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本文介绍了一种经典的二叉树遍历方法——前序遍历，并提供了递归及迭代两种实现方式。递归方法直观简洁，而迭代方法则通过使用栈来模拟递归过程，适用于对内存调用栈有限制的情况。

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[LeetCode]Binary Tree Preorder Traversal

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

For example:
Given binary tree {1,#,2,3},

return [1,2,3].

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

Recursive Method:

/**
 * 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> ans;
        if (!root) return ans;
        recursive(root, ans);
        return ans;

    }

    void recursive(TreeNode* root, vector<int>& ans)
    {
        ans.push_back(root->val);
        if(root->left) recursive(root->left, ans);
        if(root->right) recursive(root->right, ans);
    }


};

Recursive Method with on function:
key point: vector.insert()

/**
 * 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> ans;
        if (!root) return ans;
        ans.push_back(root->val);
        vector<int> left = preorderTraversal(root->left);
        ans.insert(ans.end(), left.begin(), left.end());
        vector<int> right = preorderTraversal(root->right);
        ans.insert(ans.end(), right.begin(), right.end());
        return ans;
    }
};

Iterative Method:
用栈模拟递归，先把根节点压栈，当栈非空时，访问栈顶结点，压入栈顶结点的右儿子跟左儿子。

/**
 * 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> ans;
        if(!root) return ans;
        stack<TreeNode*> s;
        s.push(root);

        while(!s.empty())
        {
            TreeNode* temp = s.top();
            ans.push_back(temp->val);
            s.pop();
            if(temp->right) s.push(temp->right);
            if(temp->left) s.push(temp->left);
        }
    }
};