662. Maximum Width of Binary Tree

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Problem

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 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

Code

如果节点 x 的 index 记为 i, 那么它的两个子节点的 index 是 2i 和 2i + 1

class Solution {
    class DataNode {
        TreeNode node;
        int index;

        public DataNode(TreeNode node, int index) {
            this.node = node;
            this.index = index;
        }
    }

    public int widthOfBinaryTree(TreeNode root) {
        if (root == null) return 0;

        Queue<DataNode> queue = new LinkedList<>();
        int maxWidth = 0;

        queue.offer(new DataNode(root, 0));

        while (!queue.isEmpty()) {
            int size = queue.size();
            int headIndex = -1;
            int tailIndex = -1;

            for (int i = 0; i < size; i++) {
                DataNode dataNode = queue.poll();
                TreeNode node = dataNode.node;
                int index = dataNode.index;

                if (i == 0) {
                    headIndex = index;
                }

                tailIndex = index;

                if (node.left != null) {
                    queue.offer(new DataNode(node.left, 2 * index));
                }

                if (node.right != null) {
                    queue.offer(new DataNode(node.right, 2 * index + 1));
                }
            }

            maxWidth = Math.max(maxWidth, tailIndex - headIndex + 1);
        }

        return maxWidth;
    }
}