LC.P662[二叉树最大宽度]

方法一：BFS

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/**
 * 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) {
        int ans = 0;
        Deque<TreeNode> queue = new ArrayDeque<>();
        // 将节点的索引值存入 val 中
        queue.offer(new TreeNode(1, root.left, root.right));
        while (!queue.isEmpty()) {
            int size = queue.size(), startIndex = -1, endIndex = -1;
            while (size-- > 0) {
                TreeNode node = queue.poll();
                endIndex = node.val;
                if (startIndex == -1) startIndex = node.val;
                if (node.left != null) {
                    queue.offer(new TreeNode(node.val * 2, node.left.left, node.left.right));
                }
                if (node.right != null) {
                    queue.offer(new TreeNode(node.val * 2 + 1, node.right.left, node.right.right));
                }
            }
            ans = Math.max(ans, endIndex - startIndex + 1);
        }
        return ans;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$

方法二：DFS

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/**
 * 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 {
    int ans = 0;
    Map<Integer, Integer> minIndex = new HashMap<>();

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

    private void dfs(TreeNode node, int index, int depth) {
        if (node == null) return;
        minIndex.putIfAbsent(depth, index);
        ans = Math.max(ans, index - minIndex.get(depth) + 1);
        dfs(node.left, index * 2, depth + 1);
        dfs(node.right, index * 2 + 1, depth + 1);
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$