662 Maximum Width of Binary Tree 二叉树最大宽度
Description:
Given the root of a binary tree, return the maximum width of the given tree.
The maximum width of a tree is the maximum width among all levels.
The width of one level is defined as the length between the end-nodes (the leftmost and rightmost non-null nodes), where the null nodes between the end-nodes are also counted into the length calculation.
It is guaranteed that the answer will in the range of 32-bit signed integer.
Example:
Example 1:
Input: root = [1,3,2,5,3,null,9]
Output: 4
Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).
Example 2:
Input: root = [1,3,null,5,3]
Output: 2
Explanation: The maximum width existing in the third level with the length 2 (5,3).
Example 3:
Input: root = [1,3,2,5]
Output: 2
Explanation: The maximum width existing in the second level with the length 2 (3,2).
Example 4:
Input: root = [1,3,2,5,null,null,9,6,null,null,7]
Output: 8
Explanation: The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).
Constraints:
The number of nodes in the tree is in the range [1, 3000].
-100 <= Node.val <= 100
题目描述:
给定一个二叉树，编写一个函数来获取这个树的最大宽度。树的宽度是所有层中的最大宽度。这个二叉树与满二叉树（full binary tree）结构相同，但一些节点为空。
每一层的宽度被定义为两个端点（该层最左和最右的非空节点，两端点间的null节点也计入长度）之间的长度。
示例 :
示例 1:
输入:
1
/ \
3 2
/ \ \
5 3 9
输出: 4
解释: 最大值出现在树的第 3 层，宽度为 4 (5,3,null,9)。
示例 2:
输入:
1
/
3
/ \
5 3
输出: 2
解释: 最大值出现在树的第 3 层，宽度为 2 (5,3)。
示例 3:
输入:
1
/ \
3 2
/
5
输出: 2
解释: 最大值出现在树的第 2 层，宽度为 2 (3,2)。
示例 4:
输入:
1
/ \
3 2
/ \
5 9
/ \
6 7
输出: 8
解释: 最大值出现在树的第 4 层，宽度为 8 (6,null,null,null,null,null,null,7)。
注意:
答案在32位有符号整数的表示范围内。
思路:
广度优先遍历
整体上还是层序遍历的模板
需要按照完全二叉树记录节点的下标
为了防止溢出, 可以记录节点下标相对于该层第一个节点的下标
时间复杂度 O(n), 空间复杂度 O(n)
代码:
C++:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution
{
public:
int widthOfBinaryTree(TreeNode* root)
{
queue> q;
q.push({root, 0});
int left = 0, right = 0, result = 0;
while (!q.empty())
{
int s = q.size(), left = q.front().second, right = q.back().second;
for (int i = 0; i < s; i++)
{
auto cur = q.front();
q.pop();
if (cur.first -> left) q.push({cur.first -> left, (cur.second << 1) - (left << 1)});
if (cur.first -> right) q.push({cur.first -> right, (cur.second << 1) + 1 - (left << 1)});
}
result = max(result, right - left + 1);
}
return result;
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public int widthOfBinaryTree(TreeNode root) {
Queue queue = new LinkedList();
queue.add(new Node(root, 0));
int left = 0, right = 0, result = 0;
while (!queue.isEmpty()) {
int s = queue.size();
for (int i = 0; i < s; i++) {
Node cur = queue.poll();
if (i == 0) left = cur.pos;
if (i == s - 1) right = cur.pos;
if (cur.node.left != null) queue.offer(new Node(cur.node.left, cur.pos * 2));
if (cur.node.right != null) queue.offer(new Node(cur.node.right, cur.pos * 2 + 1));
}
result = Math.max(result, right - left + 1);
}
return result;
}
}
class Node {
TreeNode node;
int pos;
Node(TreeNode n, int p) {
node = n;
pos = p;
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def widthOfBinaryTree(self, root: TreeNode) -> int:
result, queue = 0, [(root, 0)]
while queue:
s, result = len(queue), max(result, queue[-1][1] - queue[0][1] + 1)
for _ in range(s):
cur = queue.pop(0)
(cur[0].left and queue.append((cur[0].left, cur[1] * 2))) or (cur[0].right and queue.append((cur[0].right, cur[1] * 2 + 1)))
return result