Leetcode 285. Inorder Successor in BST (Medium) (cpp)

最新推荐文章于 2022-01-20 12:12:47 发布

Niko_Ke

最新推荐文章于 2022-01-20 12:12:47 发布

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分类专栏： Leetcode C++ C++ Leetcode Tree 文章标签： leetcode cpp

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本文介绍了一个中等难度的LeetCode题目：在给定的二叉搜索树中寻找指定节点的中序遍历后继节点。提供了一种使用栈来辅助寻找后继节点的方法，并详细阐述了算法的具体实现。

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Leetcode 285. Inorder Successor in BST (Medium) (cpp)

Tag: Tree

Difficulty: Medium

/*

285. Inorder Successor in BST (Medium)

Given a binary search tree and a node in it, find the in-order successor of that node in the BST.

Note: If the given node has no in-order successor in the tree, return null.

*/
/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
	TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
		stack<TreeNode*> s;
		iniSucSta(root, p, s);
		if (!s.empty() && s.top() == p) {
			getNextSuccessor(s);
		}
		return getNextSuccessor(s);
	}
private:
	void iniSucSta(TreeNode* root, TreeNode* p, stack<TreeNode*>& s) {
		while (root != NULL) {
			if (root->val > p->val) {
				s.push(root);
				root = root->left;
			}
			else if (root->val < p->val) {
				root = root->right;
			}
			else {
				s.push(root);
				break;
			}
		}
	}
	TreeNode* getNextSuccessor(stack<TreeNode*>& s) {
		if (s.empty()) {
			return NULL;
		}
		TreeNode* cur = s.top();
		s.pop();
		TreeNode* temp = cur;
		cur = cur->right;
		while (cur != NULL) {
			s.push(cur);
			cur = cur->left;
		}
		return temp;
	}
};