【LeetCode】144. Preorder Traversal of Binary Tree

题目

给你二叉树的根节点 root ,返回它节点值的 前序 遍历.

示例 1：

输入：root = [1,null,2,3]
输出：[1,2,3]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [1,2,3,4,5,6,7]
输出：[1,2,4,5,3,6,7]

提示：

树中节点数目在范围 [0, 100] 内
-100 <= Node.val <= 100

进阶：递归算法很简单,你可以通过迭代算法完成吗？

题解

深度优先遍历

使用迭代法

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */
class Solution {
    
public:
    vector<int> preorderTraversal(TreeNode* root) {
    
        if(root==nullptr)
            return {
    };
        vector<int> result;
        stack<TreeNode*> mystack;
        mystack.push(root);

        while(!mystack.empty())
        {
    
            TreeNode* tmp = mystack.top();
            mystack.pop();
            result.emplace_back(tmp->val);
            if(tmp->right)
                mystack.emplace(tmp->right);
            if(tmp->left)
                mystack.emplace(tmp->left);
        }
        return result;
    }
};

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */
class Solution {
    
public:
    vector<int> preorderTraversal(TreeNode* root) {
    
        if(root==nullptr)
            return {
    };
        vector<int> result;
        stack<TreeNode*> mystack;
        TreeNode* node = root;

        while(!mystack.empty() || node!=nullptr)
        {
    
            while(node!=nullptr)
            {
    
                result.emplace_back(node->val);
                mystack.emplace(node);
                node = node->left;
            }
            node = mystack.top();
            mystack.pop();
            node = node->right;
        }

        return result;
    }
};

使用递归的方法

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */
class Solution {
    
public:
    void fun(TreeNode* node,vector<int> &result)
    {
    
        result.emplace_back(node->val);
        if(node->left)
            fun(node->left,result);
        if(node->right)
            fun(node->right,result);
    }
    vector<int> preorderTraversal(TreeNode* root) {
    
        vector<int> result;
        if(root)
            fun(root,result);
        return result;
    }
};