[Locked] Closest Binary Search Tree Value & Closest Binary Search Tree Value II
Closest Binary Search Tree Value
Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target.
Note:
- Given target value is a floating point.
- You are guaranteed to have only one unique value in the BST that is closest to the target.
分析:
按照正常的搜索路径,直到搜索到叶节点,选出在这个路径上离target最近的值返回
代码:
void dfs(TreeNode *cur, int &cv, double target) {
if(!cur)
return;
double num = double(cur->val) - target;
if(abs(num) < abs(double(cv) - target))
cv = cur->val;
if(num > )
dfs(cur->left, cv, target);
else
dfs(cur->right, cv, target);
return;
}
Closest Binary Search Tree Value II
Given a non-empty binary search tree and a target value, find k values in the BST that are closest to the target.
Note:
- Given target value is a floating point.
- You may assume k is always valid, that is: k ≤ total nodes.
- You are guaranteed to have only one unique set of k values in the BST that are closest to the target.
Follow up:
Assume that the BST is balanced, could you solve it in less than O(n) runtime (where n = total nodes)?
Hint:
- Consider implement these two helper functions:
getPredecessor(N)
, which returns the next smaller node to N.getSuccessor(N)
, which returns the next larger node to N.
- Try to assume that each node has a parent pointer, it makes the problem much easier.
- Without parent pointer we just need to keep track of the path from the root to the current node using a stack.
- You would need two stacks to track the path in finding predecessor and successor node separately.
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