222. 完全二叉树的节点个数
222. 完全二叉树的节点个数
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题目
Given the root
of a complete binary tree, return the number of the nodes in the tree.
According to Wikipedia , 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 2h
nodes inclusive at the last level h
.
Design an algorithm that runs in less than O(n)
time complexity.
Example 1:
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 * 10^4]
. 0 <= Node.val <= 5 * 10^4
- The tree is guaranteed to be complete.
题目大意
输出一颗完全二叉树的结点个数。
解题思路
思路一:层序遍历
按照层序遍历一次树,然后把每一层的结点个数相加即可。
思路二:递归
关键思想是比较左右子树的深度。如果它们相等,左子树是一个满二叉树,节点总数可以直接计算 (2 ^ i - 1
)。如果深度不相等,右子树是一个满二叉树,节点总数也可以根据深度计算。递归调用处理两个子树,直到达到基本情况(一个空节点)。
countNodes
函数以完全二叉树的根节点作为输入,递归计算节点的数量。getDepth
函数通过遍历左孩子直到叶子节点,计算给定节点的深度。
代码
/**
* @param {TreeNode} root
* @return {number}
*/
var countNodes = function (root) {
let res = 0;
if (!root) return res;
let queue = [root];
while (queue.length) {
let len = queue.length;
for (let i = 0; i < len; i++) {
if (queue[i].left) queue.push(queue[i].left);
if (queue[i].right) queue.push(queue[i].right);
res++;
}
queue = queue.slice(len);
}
return res;
};
/**
* @param {TreeNode} root
* @return {number}
*/
var countNodes = function (root) {
if (!root) return 0;
const getDepth = (node) => {
let depth = 0;
while (node) {
node = node.left;
depth++;
}
return depth;
};
let leftDepth = getDepth(root.left);
let rightDepth = getDepth(root.right);
if (leftDepth == rightDepth)
return Math.pow(2, leftDepth) + countNodes(root.right);
return Math.pow(2, rightDepth) + countNodes(root.left);
};
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