g0201_0300.s0222_count_complete_tree_nodes.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0201_0300.s0222_count_complete_tree_nodes;

// #Medium #Depth_First_Search #Tree #Binary_Search #Binary_Tree #Binary_Search_II_Day_10
// #2022_07_04_Time_0_ms_(100.00%)_Space_50_MB_(37.43%)

import com_github_leetcode.TreeNode;

/*
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
/**
 * 222 - Count Complete Tree Nodes\.
 *
 * Medium
 *
 * Given the `root` of a **complete** binary tree, return the number of the nodes in the tree.
 *
 * According to **[Wikipedia](http://en.wikipedia.org/wiki/Binary_tree#Types_of_binary_trees)** , every level, except possibly the last, is completely filled in a complete binary tree, and all nodes in the last level are as far left as possible. It can have between `1` and 2^h nodes inclusive at the last level `h`.
 *
 * Design an algorithm that runs in less than `O(n)` time complexity.
 *
 * **Example 1:**
 *
 * ![](https://assets.leetcode.com/uploads/2021/01/14/complete.jpg)
 *
 * **Input:** root = [1,2,3,4,5,6]
 *
 * **Output:** 6 
 *
 * **Example 2:**
 *
 * **Input:** root = []
 *
 * **Output:** 0 
 *
 * **Example 3:**
 *
 * **Input:** root = [1]
 *
 * **Output:** 1 
 *
 * **Constraints:**
 *
 * *   The number of nodes in the tree is in the range [0, 5 * 10⁴].
 * *   0 <= Node.val <= 5 * 10⁴
 * *   The tree is guaranteed to be **complete**.
**/
public class Solution {
    public int countNodes(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftHeight = leftHeight(root);
        int rightHeight = rightHeight(root);
        // case 1: When Height(Left sub-tree) = Height(right sub-tree) 2^h - 1
        if (leftHeight == rightHeight) {
            return (1 << leftHeight) - 1;
        } else {
            return 1 + countNodes(root.left) + countNodes(root.right);
        }
    }

    private int leftHeight(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return 1 + leftHeight(root.left);
    }

    private int rightHeight(TreeNode root) {
        if (root == null) {
            return 0;
        }
        return 1 + rightHeight(root.right);
    }
}