Leetcode - 285.Inorder Successor in BST

最新推荐文章于 2022-03-22 14:01:27 发布

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最新推荐文章于 2022-03-22 14:01:27 发布

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分类专栏： leetcode 文章标签： leetcode

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本文介绍了如何在一个给定的二叉搜索树中找到特定节点的中序后继节点。提供了两种方法：一种适用于具有右子树的节点，另一种适用于没有右子树的节点，并通过代码实现了解决方案。

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Problem Description:
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.

Analysis :
The easier one: p has right subtree, then its successor is just the leftmost child of its right subtree;
The harder one: p has no right subtree, then a traversal is needed to find its successor.
For harder one :
We search from the root, each time we meet a node which val is greater than p -> val, we know that this node maybe its successor, So we record this node in suc. Then try to find the node in next level.
if (p->val < root -> val) root = root -> left;
else (p -> val > root -> val) root = root -> right;
Once we find p -> val == root -> val, we know we’ve reached at p and the current suc is just its successor.

Code:

class Solution {
public:
    TreeNode* inorderSuccessor(TreeNode* root, TreeNode* p) {
        if (p -> right) return leftMost(p -> right);
        TreeNode* suc = NULL;
        while (root) {
            if (p -> val < root -> val) {
                suc = root;
                root = root -> left;
            }
            else if (p -> val > root -> val)
                root = root -> right; 
            else break;
        }
        return suc;
    }
private:
    TreeNode* leftMost(TreeNode* node) {
        while (node -> left) node = node -> left;
        return node;
    }
};