leetcode 222: Count Complete Tree Nodes

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Count Complete Tree Nodes

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

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 2^h nodes inclusive at the last level h.

[思路]

用暴力法, recursive求会超时 O(N).   如果从某节点一直向左的高度 = 一直向右的高度, 那么以该节点为root的子树一定是complete binary tree. 而 complete binary tree的节点数,可以用公式算出 2^h - 1. 如果高度不相等, 则递归调用 return countNode(left) + countNode(right) + 1.  复杂度为O(h^2)  

[CODE]

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int countNodes(TreeNode root) {
        if(root==null) return 0;
        
        int l = getLeft(root) + 1;
        int r = getRight(root) + 1;
        
        if(l==r) {
            return (2<<(l-1)) - 1;
        } else {
            return countNodes(root.left) + countNodes(root.right) + 1;
        }
    }
    
    private int getLeft(TreeNode root) {
        int count = 0;
        while(root.left!=null) {
            root = root.left;
            ++count;
        }
        return count;
    }
    
    private int getRight(TreeNode root) {
        int count = 0;
        while(root.right!=null) {
            root = root.right;
            ++count;
        }
        return count;
    }
}

leetcode 222: Count Complete Tree Nodes

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原文地址：http://blog.csdn.net/xudli/article/details/46385011