LeetCode-Count Complete Tree Nodes

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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 2hnodes inclusive at the last level h.

也就是这个要求要比O(n)还要低，所以自然就想起来binary search了，而二叉树本身就非常适合binary search。so 肯定用二分。

而到底怎么做呢？我们可以根据数的高度来跟左右子树比，这样就可以判处不完整的树到底在那一部分。看代码，非递归版的

	public int countNodes(TreeNode root) {
		int cnt = 0;
		
		int height = height(root);
		while (root != null) {
			if (height(root.right) == height-1) {
				cnt += 1 << height-1;
				root = root.right;
			} else {
				cnt += 1 << height-2;
				root = root.left;
			}
			height--;
		}
		return cnt;
	}
	
	private int height(TreeNode root) {
		int cnt = 0;
		while (root != null) {
			cnt++;
			root = root.left;
		}
		return cnt;
	}

    int height(TreeNode root) {
        return root == null ? -1 : 1 + height(root.left);
    }
    public int countNodes(TreeNode root) {
        int h = height(root);
        return h < 0 ? 0 :
               height(root.right) == h-1 ? (1 << h) + countNodes(root.right)
                                         : (1 << h-1) + countNodes(root.left);
    }

LeetCode-Count Complete Tree Nodes

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原文地址：http://blog.csdn.net/my_jobs/article/details/48104161