【LeetCode】107. 二叉树的层次遍历 II

给定一个二叉树,返回其节点值自底向上的层次遍历。 (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)

解题思路:利用之前二叉树的层次遍历的方法,在temp压入result之前,先用栈保存temp,实际上是完成一个逆序的过程。

vector<vector<int>> Solution::levelOrderBottom(TreeNode* root)
{
    vector<vector<int>> result;
    vector<int> temp;
    queue<TreeNode*> TreeQueue;
    stack<vector<int>> TreeStack;
    unsigned long cursize = 0;
    
    if(root == NULL)
    {
        return result;
    }
    TreeQueue.push(root);
    while(TreeQueue.size() != 0)
    {
        cursize = TreeQueue.size();
        for(unsigned int index = 0; index < cursize; index ++)
        {
            TreeNode* indexnode = TreeQueue.front();
            TreeQueue.pop();
            temp.push_back(indexnode->val);
            if(indexnode->left != NULL)
            {
                TreeQueue.push(indexnode->left);
            }
            if(indexnode->right != NULL)
            {
                TreeQueue.push(indexnode->right);
            }
        }
        TreeStack.push(temp);
        temp.clear();
    }
    while(TreeStack.empty() != true)
    {
        result.push_back(TreeStack.top());
        TreeStack.pop();
    }
    return result;
}

猜你喜欢

转载自blog.csdn.net/syc233588377/article/details/86626928