p32 二叉树的前序遍历 (leetcode 144)

Posted 2020-11-26 repinkply

一：解题思路

这道题目2种做法。第一种做法就是递归法，第二种就是迭代法。这2种方法的时间复杂度和空间复杂度都为O(n)。

二：完整代码示例 (C++版和Java版)

递归C++:

class Solution 
{
public:
    void preorder(TreeNode* root, vector<int>& ret)
    {
        if (root != NULL)
        {
            ret.push_back(root->val);
            preorder(root->left,ret);
            preorder(root->right,ret);
        }
    }

    vector<int> preorderTraversal(TreeNode* root) 
    {
        vector<int> ret;
        
        preorder(root,ret);

        return ret;
    }
};

递归Java:

class Solution {
    public void preorder(TreeNode root,List<Integer> ret)
    {
        if(root!=null)
        {
            ret.add(root.val);
            preorder(root.left,ret);
            preorder(root.right,ret);
        }
    }
    
    public List<Integer> preorderTraversal(TreeNode root)
    {
          List<Integer> ret=new ArrayList<>();
          
          preorder(root,ret);
          
          return ret;
    }
}

迭代C++:

class Solution 
{
public:
    vector<int> preorderTraversal(TreeNode* root) 
    {
        vector<int> result;
        stack<TreeNode*> stack;

        while (root != NULL || !stack.empty())
        {
            while (root != NULL)
            {
                result.push_back(root->val);
                stack.push(root);
                root = root->left;
            }

            root=stack.top()->right;
            stack.pop();
        }

        return result;
    }
};

迭代Java:

class Solution {
    public List<Integer> preorderTraversal(TreeNode root)
    {
          List<Integer> ret=new ArrayList<>();
          Stack<TreeNode> stack=new Stack<>();
          
          while(root!=null || !stack.isEmpty())
          {
              while(root!=null)
              {
                  ret.add(root.val);
                  stack.push(root);
                  root=root.left;
              }
              
              root=stack.pop().right;
          }

          return ret;
    }
}