Formatted question description: https://leetcode.ca/all/662.html

662. Maximum Width of Binary Tree (Medium)

Given a binary tree, write a function to get the maximum width of the given tree. The width of a tree is the maximum width among all levels. The binary tree has the same structure as a full binary tree, but some nodes are null.

The width of one level is defined as the length between the end-nodes (the leftmost and right most non-null nodes in the level, where the null nodes between the end-nodes are also counted into the length calculation.

Example 1:

Input:

1
/   \
3     2
/ \     \
5   3     9

Output: 4
Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).


Example 2:

Input:

1
/
3
/ \
5   3

Output: 2
Explanation: The maximum width existing in the third level with the length 2 (5,3).


Example 3:

Input:

1
/ \
3   2
/
5

Output: 2
Explanation: The maximum width existing in the second level with the length 2 (3,2).


Example 4:

Input:

1
/ \
3   2
/     \
5       9
/         \
6           7
Output: 8
Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).



Note: Answer will in the range of 32-bit signed integer.

Companies:
Bloomberg, Microsoft

Related Topics:
Tree

Solution 1. Pre-order Traversal

// OJ: https://leetcode.com/problems/maximum-width-of-binary-tree/
// Time: O(N)
// Space: O(H)
class Solution {
typedef unsigned long long ULL;
vector<vector<ULL>> v;
ULL ans = 0;
void dfs(TreeNode *node, int lv, ULL nodeId) {
if (!node) return;
if (v.size() <= lv) v.push_back({ULLONG_MAX, 0});
v[lv][0] = min(v[lv][0], nodeId);
v[lv][1] = max(v[lv][1], nodeId);
ans = max(ans, v[lv][1] - v[lv][0] + 1);
dfs(node->left, lv + 1, 2 * nodeId);
dfs(node->right, lv + 1, 2 * nodeId + 1);
}
public:
int widthOfBinaryTree(TreeNode* root) {
if (!root) return 0;
dfs(root, 0, 0);
return ans;
}
};


Solution 1. Level-order Traversal

// OJ: https://leetcode.com/problems/maximum-width-of-binary-tree/
// Time: O(N)
// Space: O(N)
class Solution {
typedef unsigned long long ULL;
public:
int widthOfBinaryTree(TreeNode* root) {
if (!root) return 0;
queue<pair<TreeNode*, ULL>> q;
q.emplace(root, 0);
ULL ans = 0;
while (q.size()) {
ULL cnt = q.size(), minId, maxId;
for (int i = 0; i < cnt; ++i) {
auto p = q.front();
q.pop();
auto node = p.first;
ULL nodeId = p.second;
if (node->left) q.emplace(node->left, nodeId * 2);
if (node->right) q.emplace(node->right, nodeId * 2 + 1);
if (i == 0) minId = nodeId;
if (i == cnt - 1) maxId = nodeId;
}
ans = max(ans, maxId - minId + 1);
}
return ans;
}
};


Solution 3. Level-order Traversal

When there is only one node in this level, reset the id to be 0.

// OJ: https://leetcode.com/problems/maximum-width-of-binary-tree/
// Time: O(N)
// Space: O(N)
class Solution {
public:
int widthOfBinaryTree(TreeNode* root) {
if (!root) return 0;
queue<pair<TreeNode*, int>> q;
q.emplace(root, 0);
int ans = 0;
while (q.size()) {
int cnt = q.size();
if (cnt == 1) {
q.emplace(q.front().first, 0);
q.pop();
}
ans = max(ans, q.back().second - q.front().second + 1);
while (cnt--) {
root = q.front().first;
int id = q.front().second;
q.pop();
if (root->left) q.emplace(root->left, 2 * id);
if (root->right) q.emplace(root->right, 2 * id + 1);
}
}
return ans;
}
};

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int widthOfBinaryTree(TreeNode root) {
if (root == null)
return 0;
int maxWidth = 0;
nodeQueue.offer(root);
numQueue.offer(0);
while (!nodeQueue.isEmpty()) {
int leftmost = -1, rightmost = -1;
int size = nodeQueue.size();
for (int i = 0; i < size; i++) {
TreeNode node = nodeQueue.poll();
int num = numQueue.poll();
if (leftmost < 0)
leftmost = num;
rightmost = num;
TreeNode left = node.left, right = node.right;
if (left != null) {
nodeQueue.offer(left);
numQueue.offer(num * 2 + 1);
}
if (right != null) {
nodeQueue.offer(right);
numQueue.offer(num * 2 + 2);
}
}
int width = rightmost - leftmost + 1;
maxWidth = Math.max(maxWidth, width);
}
return maxWidth;
}
}

############

/**
* 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) {
Deque<Pair<TreeNode, Integer>> q = new ArrayDeque<>();
q.offer(new Pair<>(root, 1));
int ans = 0;
while (!q.isEmpty()) {
ans = Math.max(ans, q.peekLast().getValue() - q.peekFirst().getValue() + 1);
for (int n = q.size(); n > 0; --n) {
var p = q.pollFirst();
root = p.getKey();
int i = p.getValue();
if (root.left != null) {
q.offer(new Pair<>(root.left, i << 1));
}
if (root.right != null) {
q.offer(new Pair<>(root.right, i << 1 | 1));
}
}
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/maximum-width-of-binary-tree/
// Time: O(N)
// Space: O(H)
class Solution {
typedef unsigned long long ULL;
vector<vector<ULL>> v;
ULL ans = 0;
void dfs(TreeNode *node, int lv, ULL nodeId) {
if (!node) return;
if (v.size() <= lv) v.push_back({ULLONG_MAX, 0});
v[lv][0] = min(v[lv][0], nodeId);
v[lv][1] = max(v[lv][1], nodeId);
ans = max(ans, v[lv][1] - v[lv][0] + 1);
dfs(node->left, lv + 1, 2 * nodeId);
dfs(node->right, lv + 1, 2 * nodeId + 1);
}
public:
int widthOfBinaryTree(TreeNode* root) {
if (!root) return 0;
dfs(root, 0, 0);
return ans;
}
};

• # 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: Optional[TreeNode]) -> int:
ans = 0
q = deque([(root, 1)])
while q:
ans = max(ans, q[-1][1] - q[0][1] + 1)
for _ in range(len(q)):
root, i = q.popleft()
if root.left:
q.append((root.left, i << 1))
if root.right:
q.append((root.right, i << 1 | 1))
return ans

############

class Solution(object):
def widthOfBinaryTree(self, root):
"""
:type root: TreeNode
:rtype: int
"""

def dfs(root, x, y, num, dmin, dmax):
if root:
left = dfs(root.left, x - 1, y + 1, num * 2, dmin, dmax)
right = dfs(root.right, x + 1, y + 1, 1 + num * 2, dmin, dmax)
dmin[y] = min(num, dmin.get(y, float("inf")))
dmax[y] = max(num, dmax.get(y, float("-inf")))
return max(left or 0, right or 0, 1 + dmax[y] - dmin[y])

return dfs(root, 0, 0, 1, {}, {})


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func widthOfBinaryTree(root *TreeNode) int {
q := []pair{ {root, 1} }
ans := 0
for len(q) > 0 {
ans = max(ans, q[len(q)-1].i-q[0].i+1)
for n := len(q); n > 0; n-- {
p := q[0]
q = q[1:]
root = p.node
if root.Left != nil {
q = append(q, pair{root.Left, p.i << 1})
}
if root.Right != nil {
q = append(q, pair{root.Right, p.i<<1 | 1})
}
}
}
return ans
}

type pair struct {
node *TreeNode
i    int
}

func max(a, b int) int {
if a > b {
return a
}
return b
}