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662. Maximum Width of Binary Tree

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

BFS or DFS.

• Java
• C++
• Python
• Go

• /**
   * 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.
  # 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.
   * 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
  }
  

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