222.Count-Complete-Tree-Nodes.md

222. Count Complete Tree Nodes

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题目地址

https://leetcode.com/problems/count-complete-tree-nodes/

题目描述

Given a complete binary tree, count the number of nodes.

Note:

Definition of a complete binary tree from Wikipedia:
In a complete binary tree every level, except possibly the last, is completely filled, and all nodes in the last level are as far left as possible. It can have between 1 and 2h nodes inclusive at the last level h.

Example:

Input: 
    1
   / \
  2   3
 / \  /
4  5 6

Output: 6

代码

Approach #1 Linear Time

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
  public int countNodes(TreeNode root) {
		return root != null ? 1 + countNodes(root.right) + countNodes(left) : 0;
  }
}

Approach #2 Binary Search

class Solution {
  // Last level nodes are enumerated from 0 to 2**d - 1 (left -> right).
  // Return True if last level node idx exists. 
  // Binary search with O(d) complexity.
  public boolean exists(int idx, int d, TreeNode node) {
    int left = 0, right = (int)Math.pow(2, d) - 1;
    int pivot;
    for(int i = 0; i < d; ++i) {
      pivot = left + (right - left) / 2;
      if (idx <= pivot) {
        node = node.left;
        right = pivot;
      }
      else {
        node = node.right;
        left = pivot + 1;
      }
    }
    return node != null;
  }

  public int countNodes(TreeNode root) {
    // if the tree is empty
    if (root == null) return 0;

    int d = computeDepth(root);
    // if the tree contains 1 node
    if (d == 0) return 1;

    // Last level nodes are enumerated from 0 to 2**d - 1 (left -> right).
    // Perform binary search to check how many nodes exist.
    int left = 1, right = (int)Math.pow(2, d) - 1;
    int pivot;
    while (left <= right) {
      pivot = left + (right - left) / 2;
      if (exists(pivot, d, root)) left = pivot + 1;
      else right = pivot - 1;
    }

    // The tree contains 2**d - 1 nodes on the first (d - 1) levels
    // and left nodes on the last level.
    return (int)Math.pow(2, d) - 1 + left;
  }
  
    // Return tree depth in O(d) time.
  public int computeDepth(TreeNode node) {
    int d = 0;
    while (node.left != null) {
      node = node.left;
      ++d;
    }
    return d;
  }
}