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面试题 04.06. 后继者

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Symbols count in article: 1.4k Reading time ≈ 1 mins.

面试题 04.06. 后继者

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/**
 * 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) {
        TreeNode* pre = nullptr;
        TreeNode* cur = root;
        stack<TreeNode*> s;
        while(!s.empty() || cur)
        {
            while(cur)
            {
                s.push(cur);
                cur = cur->left;
            }
            cur = s.top();
            s.pop();
            if(pre == p)
                return cur;
            pre = cur;
            cur = cur->right;
        }
        return nullptr;
    }
};
利用bst性质

后继节点是大于他中的最小的数

如果这个节点有右子树，后继节点就是右子树的最左边，如果没有，后继节点就是他路上最小的一个。

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/**
 * 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) {
        TreeNode* cur = root;
        TreeNode* ans = nullptr;
        while(cur)
        {
            if(cur->val > p->val && (!ans || ans->val > cur->val))
                ans = cur; // 记录路上大于p中最小的点
            if(cur == p)
            {
                if(cur->right) // 如果有右数，就返回右数中的最左边
                {
                    cur = cur->right;
                    while(cur->left)
                        cur = cur->left;
                    return cur;
                }else // 反之，返回路上的点
                    return ans;
            }else
            {
                if(cur->val > p->val)
                    cur = cur->left;
                else
                    cur = cur->right;
            }
        }
        return nullptr;
    }
};

O(logn)