力扣 173.二叉搜索树迭代器

题面

题解(二叉树的非递归中序遍历)

将整棵树的最左边的一条链压入栈中，每次取出栈顶元素，并记录，如果它有右子树，那么将右子树最左边压入压栈中

代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class BSTIterator {
public:

    stack<TreeNode*> stk; 

    BSTIterator(TreeNode* root) {
        while(root){
            stk.push(root);
            root=root->left;
        }
    }
    
    int next() {
        auto root =stk.top();
        stk.pop();
        int val = root -> val;
        root = root -> right;
        while(root){
            stk.push(root);
            root = root -> left;
        }
        return val; 
    }
    
    bool hasNext() {
        return stk.size();
    }
};

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator* obj = new BSTIterator(root);
 * int param_1 = obj->next();
 * bool param_2 = obj->hasNext();
 */

标签：right,TreeNode,val,int,173,二叉,力扣,root,left
来源： https://blog.csdn.net/qq_44791484/article/details/115791279