LeetCode – Refresh – Binary Tree Iterator

最新推荐文章于 2022-07-08 20:27:39 发布

转载最新推荐文章于 2022-07-08 20:27:39 发布 · 64 阅读

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原文链接：http://www.cnblogs.com/shuashuashua/p/4346167.html

本文介绍了一种使用栈来实现二叉搜索树(BST)中序遍历的方法，通过不断将右子节点压入栈中，并在弹出栈顶元素后再次将其右子树的所有左子节点压入栈中，以此确保每次都能获取到下一个最小值。

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It is similar to use stack for BST inorder traversal.

So do the same thing :

1. Get stack to store right branchs (include current node).

2. When you pop the top node, remember to push all the right sub branches again. Otherwise, it will cause right's left branches missed from result

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class BSTIterator {
11 private:
12     stack<TreeNode *> s;
13 public:
14     void initialNodes(TreeNode *root) {
15         while (root) {
16             s.push(root);
17             root = root->left;
18         }
19     }
20     BSTIterator(TreeNode *root) {
21         initialNodes(root);
22     }
23 
24     /** @return whether we have a next smallest number */
25     bool hasNext() {
26         return !s.empty();
27     }
28 
29     /** @return the next smallest number */
30     int next() {
31         TreeNode *tmp = s.top();
32         s.pop();
33         if (tmp->right) {
34             initialNodes(tmp->right);
35         }
36         return tmp->val;
37     }
38 };
39 
40 /**
41  * Your BSTIterator will be called like this:
42  * BSTIterator i = BSTIterator(root);
43  * while (i.hasNext()) cout << i.next();
44  */

转载于:https://www.cnblogs.com/shuashuashua/p/4346167.html