Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST.
Assume a BST is defined as follows:
The left subtree of a node contains only nodes with keys less than or equal to the node's key.
The right subtree of a node contains only nodes with keys greater than or equal to the node's key.
Both the left and right subtrees must also be binary search trees.
For example:
Given BST [1,null,2,2],
1
\
2
/
2
return [2].
Note: If a tree has more than one mode, you can return them in any order.
Follow up: Could you do that without using any extra space? (Assume that the implicit stack space incurred due to recursion does not count).
Solution：中序遍历
思路：
Time Complexity: O(N) Space Complexity: O(N)
Solution Code：
class Solution {
Integer prev = null;
int count = 1;
int max = 0;
public int[] findMode(TreeNode root) {
if (root == null) return new int[0];
List<Integer> list = new ArrayList<>();
traverse(root, list);
int[] res = new int[list.size()];
for (int i = 0; i < list.size(); ++i) res[i] = list.get(i);
return res;
}
private void traverse(TreeNode root, List<Integer> list) {
if (root == null) return;
traverse(root.left, list);
if (prev != null) {
if (root.val == prev)
count++;
else
count = 1;
}
if (count > max) {
max = count;
list.clear();
list.add(root.val);
} else if (count == max) {
list.add(root.val);
}
prev = root.val;
traverse(root.right, list);
}
}
