二叉树遍历（非递归

最新推荐文章于 2022-12-09 18:31:27 发布

sdsfdsfdsds

最新推荐文章于 2022-12-09 18:31:27 发布

阅读量92

点赞数

分类专栏：编程知识文章标签： stack

本文链接：https://blog.youkuaiyun.com/sdsfdsfdsds/article/details/108362634

版权

编程知识专栏收录该内容

13 篇文章

订阅专栏

前序遍历

vector<int> preorderTraversal(TreeNode* root) {
        vector<int> res;
        stack<TreeNode*> stack1;
        if(root==NULL)  return res;
        stack1.push(root);
        while(!stack1.empty())
        {
            TreeNode* topnum=stack1.top();
            stack1.pop();
            res.push_back(topnum->val);
            if(topnum->right)   
            	stack1.push(topnum->right);
            if(topnum->left)    
            	stack1.push(topnum->left);
        }
        return res;
    }

中序遍历

vector<int> inorderTraversal(TreeNode* root) {
        stack<TreeNode*> stack1;
        vector<int> res;
        if(root==NULL)  return res;
        TreeNode* cur=root;
        while(cur!=NULL||!stack1.empty())
        {
            while(cur!=NULL)
            {
                stack1.push(cur);
                cur=cur->left;
            }
            cur=stack1.top();
            stack1.pop();
            res.push_back(cur->val);
            cur=cur->right;
        }
        return res;
    }

后续遍历

前序是根左右，后续是左右根，因此先生成根右左，再反转得到左右根

 
 vector<int> postorderTraversal(TreeNode* root) {
 		vector<int> res;
        if(root==NULL)    return res;
        stack<TreeNode*> stack1;
        stack1.push(root);
        while(!stack1.empty())
        {
            TreeNode* cur=stack1.top();
            stack1.pop();
            res.push_back(cur->val);
            if(cur->left)   
            	stack1.push(cur->left);
            if(cur->right)  
            	stack1.push(cur->right);
        }
        reverse(res.begin(),res.end());
        return res;
    }