Algorithm

Binary Tree iteration

递归遍历

//前序遍历
void preorder(TreeNode *root, vector<int> &path)
{
    if(root != NULL)
    {
        path.push_back(root->val);
        preorder(root->left, path);
        preorder(root->right, path);
    }
}

//中序遍历
void inorder(TreeNode *root, vector<int> &path)
{
    if(root != NULL)
    {
        inorder(root->left, path);
        path.push_back(root->val);
        inorder(root->right, path);
    }
}

//后续遍历
void postorder(TreeNode *root, vector<int> &path)
{
    if(root != NULL)
    {
        postorder(root->left, path);
        postorder(root->right, path);
        path.push_back(root->val);
    }
}

非递归遍历 Reference: 更简单的非递归遍历二叉树的方法

非递归遍历二叉树实际上就是看是否第二次遍历，第二次遍历的时候加入到结果集合中即可，按照这个思想就可以借助stack的先进先出控制树的结点的遍历：

//更简单的非递归前序遍历
void preorderTraversalNew(TreeNode *root, vector<int> &path)
{
    stack< pair<TreeNode *, bool> > s;
    s.push(make_pair(root, false));
    bool visited;
    while(!s.empty())
    {
        root = s.top().first;
        visited = s.top().second;
        s.pop();
        if(root == NULL)
            continue;
        if(visited)
        {
            path.push_back(root->val);
        }
        else
        {
            s.push(make_pair(root->right, false));
            s.push(make_pair(root->left, false));
            s.push(make_pair(root, true));
        }
    }
}

//更简单的非递归中序遍历
void inorderTraversalNew(TreeNode *root, vector<int> &path)
{
    stack< pair<TreeNode *, bool> > s;
    s.push(make_pair(root, false));
    bool visited;
    while(!s.empty())
    {
        root = s.top().first;
        visited = s.top().second;
        s.pop();
        if(root == NULL)
            continue;
        if(visited)
        {
            path.push_back(root->val);
        }
        else
        {
            s.push(make_pair(root->right, false));
            s.push(make_pair(root, true));
            s.push(make_pair(root->left, false));
        }
    }
}

//更简单的非递归后序遍历
void postorderTraversalNew(TreeNode *root, vector<int> &path)
{
    stack< pair<TreeNode *, bool> > s;
    s.push(make_pair(root, false));
    bool visited;
    while(!s.empty())
    {
        root = s.top().first;
        visited = s.top().second;
        s.pop();
        if(root == NULL)
            continue;
        if(visited)
        {
            path.push_back(root->val);
        }
        else
        {
            s.push(make_pair(root, true));
            s.push(make_pair(root->right, false));
            s.push(make_pair(root->left, false));
        }
    }
}

DFS & BFS

https://leetcode-cn.com/problems/binary-tree-level-order-traversal/solution/bfs-de-shi-yong-chang-jing-zong-jie-ceng-xu-bian-l/