Question

662. Maximum Width of Binary Tree

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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 that would be present in a complete binary tree extending down to that level are also counted into the length calculation.

It is guaranteed that the answer will in the range of a 32-bit signed integer.

 

Example 1:

Input: root = [1,3,2,5,3,null,9]
Output: 4
Explanation: The maximum width exists in the third level with length 4 (5,3,null,9).

Example 2:

Input: root = [1,3,2,5,null,null,9,6,null,7]
Output: 7
Explanation: The maximum width exists in the fourth level with length 7 (6,null,null,null,null,null,7).

Example 3:

Input: root = [1,3,2,5]
Output: 2
Explanation: The maximum width exists in the second level with length 2 (3,2).

 

Constraints:

• The number of nodes in the tree is in the range [1, 3000].
• -100 <= Node.val <= 100

Solutions

Solution 1

# 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

/**
 * 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;
  }
}

/**
 * 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<pair<TreeNode*, int>> q;
    q.push({root, 1});
    int ans = 0;
    while (!q.empty()) {
      ans = max(ans, q.back().second - q.front().second + 1);
      int i = q.front().second;
      for (int n = q.size(); n; --n) {
        auto p = q.front();
        q.pop();
        root = p.first;
        int j = p.second;
        if (root->left) q.push({root->left, (j << 1) - (i << 1)});
        if (root->right) q.push({root->right, (j << 1 | 1) - (i << 1)});
      }
    }
    return ans;
  }
};

/**
 * 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
}

Solution 2

# 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:
    def dfs(root, depth, i):
      if root is None:
        return
      if len(t) == depth:
        t.append(i)
      else:
        nonlocal ans
        ans = max(ans, i - t[depth] + 1)
      dfs(root.left, depth + 1, i << 1)
      dfs(root.right, depth + 1, i << 1 | 1)

    ans = 1
    t = []
    dfs(root, 0, 1)
    return ans

/**
 * 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 {
  private int ans = 1;
  private List<Integer> t = new ArrayList<>();

  public int widthOfBinaryTree(TreeNode root) {
    dfs(root, 0, 1);
    return ans;
  }

  private void dfs(TreeNode root, int depth, int i) {
    if (root == null) {
      return;
    }
    if (t.size() == depth) {
      t.add(i);
    } else {
      ans = Math.max(ans, i - t.get(depth) + 1);
    }
    dfs(root.left, depth + 1, i << 1);
    dfs(root.right, depth + 1, i << 1 | 1);
  }
}

/**
 * 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:
  vector<int> t;
  int ans = 1;
  using ull = unsigned long long;

  int widthOfBinaryTree(TreeNode* root) {
    dfs(root, 0, 1);
    return ans;
  }

  void dfs(TreeNode* root, int depth, ull i) {
    if (!root) return;
    if (t.size() == depth) {
      t.push_back(i);
    } else {
      ans = max(ans, (int) (i - t[depth] + 1));
    }
    dfs(root->left, depth + 1, i << 1);
    dfs(root->right, depth + 1, i << 1 | 1);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func widthOfBinaryTree(root *TreeNode) int {
  ans := 1
  t := []int{}
  var dfs func(root *TreeNode, depth, i int)
  dfs = func(root *TreeNode, depth, i int) {
    if root == nil {
      return
    }
    if len(t) == depth {
      t = append(t, i)
    } else {
      ans = max(ans, i-t[depth]+1)
    }
    dfs(root.Left, depth+1, i<<1)
    dfs(root.Right, depth+1, i<<1|1)
  }
  dfs(root, 0, 1)
  return ans
}