LEETCODE 222. 完全二叉树的节点个数

递归

递归很简单，遍历整棵树即可，代码复杂度为O(n)

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func countNodes(root *TreeNode) int {
    if root == nil {
        return 0
    }
    return 1+countNodes(root.Left)+countNodes(root.Right)
}

二分查找

二分查找的思路如下图所示

最难的部分应该是判断某个节点是否存在，需要结合完全二叉树的性质，只要给出一个n就可以构造出从根节点到该叶子节点的路径，我们找到之后判断该节点是否为nil即可。

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func countNodes(root *TreeNode) int {
	if root == nil {
		return 0
	}
	var bigLeft, bigRight = getRange(root)
	var left, right = bigLeft, bigRight
	for left < right {
		var mid = left + (right-left)/2
		if ex(root, bigLeft, bigRight, mid) {
			left = mid + 1
		} else {
			right = mid
		}
	}
	return left - 1
}

//左闭右开
func getRange(root *TreeNode) (int, int) {
	if root == nil {
		return 0, 0
	}
	var res = 1
	var node = root
	for node.Left != nil {
		res = res * 2
		node = node.Left
	}
	return res, res * 2
}

func ex(root *TreeNode, left int, right int, n int) bool {
	if root == nil {
		return false
	}
	if root.Left == nil && root.Right == nil {
		return n == 1
	}
	var node = root
	for node != nil && left < right {
		var mid = left + (right-left)/2
		if right-left == 1 {
			node = node.Left
			break
		}
		if right-left == 2 {
			if n == left {
				node = node.Left
			} else if n == left+1 {
				node = node.Right
			}
			break
		}
		if left <= n && n < mid {
			node = node.Left
			right = mid
		} else if mid <= n && n < right {
			node = node.Right
			left = mid
		}
	}
	return node != nil
}