(nice!!!)(LeetCode 面试经典 150 题) 173. 二叉搜索树迭代器 (栈)

题目：173. 二叉搜索树迭代器

思路：栈，时间复杂度0(n)。

C++版本：

cpp 复制代码

/**
 * 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*> st;
    TreeNode * tr; 
    BSTIterator(TreeNode* root) {
        tr=root;
    }
    int next() {
    	//每次先中序遍历tr
        while(tr!=nullptr){
            st.push(tr);
            tr=tr->left;
        }
        // 确保当前是中序遍历的第一个节点
        tr=st.top();
        st.pop();
        int tmp=tr->val;
        //保留当前节点的右子树，便于下次接着进行中序遍历
        tr=tr->right;
        return tmp;
    }
    
    bool hasNext() {
        return tr!=nullptr || st.size()!=0 ;
    }
};

/**
 * 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();
 */

JAVA版本：

java 复制代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class BSTIterator {
    TreeNode tr;
    Deque<TreeNode> st;
    public BSTIterator(TreeNode root) {
        tr=root;
        st=new LinkedList<TreeNode>();
    }
    
    public int next() {
        while(tr!=null){
            st.push(tr);
            tr=tr.left;
        }
        tr=st.pop();
        int tmp=tr.val;
        tr=tr.right;
        return tmp;
    }
    
    public boolean hasNext() {
        return tr!=null || !st.isEmpty() ;
    }
}

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

GO版本：

go 复制代码

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
type BSTIterator struct {
    tr *TreeNode
    st []*TreeNode
}


func Constructor(root *TreeNode) BSTIterator {
    return BSTIterator{tr:root}
}


func (this *BSTIterator) Next() int {
    for this.tr!=nil {
        this.st=append(this.st,this.tr)
        this.tr=this.tr.Left
    }
    this.tr=this.st[len(this.st)-1];
    this.st=this.st[:len(this.st)-1]
    tmp:=this.tr.Val
    this.tr=this.tr.Right
    return tmp
}


func (this *BSTIterator) HasNext() bool {
    return this.tr!=nil || len(this.st)>0
}


/**
 * Your BSTIterator object will be instantiated and called as such:
 * obj := Constructor(root);
 * param_1 := obj.Next();
 * param_2 := obj.HasNext();
 */