[C++ code]手撕二叉树的遍历

前序遍历

根-左-右，递归

class Solution{
public:
    vector preorderTraversal(TreeNode* root)
    {
      vector result;
      preorder(root, result);
      return result;
    }
    void preorder(TreeNode* cur, vector &res)
    {
      if(cur == nullptr) return;
      res.push_back(cur->val);
      preorder(cur->left, res);
      preorder(cur->right, res);
    }

};

中序遍历

左-根-右

class Sulution{
public{
    vector inorderTranserval(TreeNode* root)
    {
      verctor result;
      inorder(root, result);
      return result;
    }
    void inorder(TreeNode* cur, vector &res)
    {
      if(cur == nullptr){return;}
      inorder(cur->left, res);
      res.push_back(cur->val);
      inorder(cur->right,res);
    }
}
};

后序遍历

class Solution {
public:
    vector postorderTraversal(TreeNode* root) {
        vector result;
        postorder(root, result);
        return result;

    }
    void postorder(TreeNode* cur, vector &res){
        if(cur == nullptr){
            return;
        }
        postorder(cur->left, res);
        postorder(cur->right, res);
        res.push_back(cur->val);
    }
};

层序遍历

入队根节点，出队根节点，入队他的左子树和右子树，出队左子树，入队左子树的左子树和右子树。。。以此类推。

class solution
{
public:
    vector> levelOrder(TreeNode* root)
    {
      vector> res;
      if(root == nullptr){return res;}
      queue q;
      q.push(root);
      while(!q.empty)
      {
        vector vec;
        int qSize = q.size();
        for(int i = 0; ival);
           if(node->left)q.push(node->left);
           if(node->right)q.push(node->right);
        }
      res.push_bac(vec);
      }
    return res;
    }
};

层序遍历Ⅱ

给你二叉树的根节点 root ，返回其节点值 自底向上的层序遍历 。 （即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

与层序遍历相同，只是在输出前反转一下即可。

reverse(res.begin(),res.end());