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

Solution 1

Python3JavaC++Go

# 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

Python3JavaC++Go

# 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
}

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