Find Mode in Binary Search Tree

Given a binary search tree (BST) with duplicates, find all the mode(s) (the most frequently occurred element) in the given BST.

Thoughts

看到BST想到它的性质用中序遍历会得到排好序的结点，然后我们就可以数相同元素有几个了。为了节省空间, 可以用两次中序遍历存储maxLen.

Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private void pushLeft(Stack<TreeNode> stack, TreeNode node) {
        while (node != null) {
            stack.push(node);
            node = node.left;
        }
    }

    public int[] findMode(TreeNode root) {
        TreeNode last = null;
        int max = 0, seqLen = 0, maxCount = 0;
        Stack<TreeNode> stack = new Stack<>();
        pushLeft(stack, root);
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            if (last != null && node.val == last.val) {
                seqLen++;
            } else {
                seqLen = 1;
            }
            if (seqLen == max) {
                maxCount++;
            } else if (seqLen > max) {
                maxCount = 1;
                max = seqLen;
            }
            last = node;
            pushLeft(stack, node.right);
        }
        last = null;
        int[] res = new int[maxCount];
        pushLeft(stack, root);
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            if (last != null && node.val == last.val) {
                seqLen++;
            } else {
                seqLen = 1;
            }
            if (seqLen == max) {
                res[--maxCount] = node.val;
            }
            last = node;
            pushLeft(stack, node.right);
        }
        return res;
    }
}

Analysis

Errors: 1. 老版本调用两次inOder时第二次忘了把last设成null 1. seqLen >= maxLen就清空了res, 导致前面的==max的被删掉

时间复杂度为中序遍历O(n).

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Last updated 6 years ago

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Code

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private void pushLeft(Stack<TreeNode> stack, TreeNode node) {
        while (node != null) {
            stack.push(node);
            node = node.left;
        }
    }

    public int[] findMode(TreeNode root) {
        TreeNode last = null;
        int max = 0, seqLen = 0, maxCount = 0;
        Stack<TreeNode> stack = new Stack<>();
        pushLeft(stack, root);
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            if (last != null && node.val == last.val) {
                seqLen++;
            } else {
                seqLen = 1;
            }
            if (seqLen == max) {
                maxCount++;
            } else if (seqLen > max) {
                maxCount = 1;
                max = seqLen;
            }
            last = node;
            pushLeft(stack, node.right);
        }
        last = null;
        int[] res = new int[maxCount];
        pushLeft(stack, root);
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            if (last != null && node.val == last.val) {
                seqLen++;
            } else {
                seqLen = 1;
            }
            if (seqLen == max) {
                res[--maxCount] = node.val;
            }
            last = node;
            pushLeft(stack, node.right);
        }
        return res;
    }
}

Analysis

Errors: 1. 老版本调用两次inOder时第二次忘了把last设成null 1. seqLen >= maxLen就清空了res, 导致前面的==max的被删掉

时间复杂度为中序遍历O(n).