LC.P1609[奇偶树]

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 boolean isEvenOddTree(TreeNode root) {
        Deque<TreeNode> queue = new ArrayDeque<>();
        queue.offer(root);
        int depth = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> list = new ArrayList<>();
            while (size-- > 0) {
                TreeNode cur = queue.poll();
                if (depth % 2 == 0) {
                    if (cur.val % 2 == 0) return false;
                    if (!list.isEmpty() && list.get(list.size() - 1) >= cur.val) return false;
                } else {
                    if (cur.val % 2 == 1) return false;
                    if (!list.isEmpty() && list.get(list.size() - 1) <= cur.val) return false;
                }
                list.add(cur.val);
                if (cur.left != null) queue.offer(cur.left);
                if (cur.right != null) queue.offer(cur.right);
            }
            ++depth;
        }
        return true;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$

方法二：优化

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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 boolean isEvenOddTree(TreeNode root) {
        Deque<TreeNode> queue = new ArrayDeque<>();
        queue.offer(root);
        boolean flag = true;
        while (!queue.isEmpty()) {
            int size = queue.size(), prev = flag ? 0 : Integer.MAX_VALUE;
            while (size-- > 0) {
                TreeNode node = queue.poll();
                int cur = node.val;
                if (flag && (cur % 2 == 0 || cur <= prev)) return false;
                if (!flag && (cur % 2 != 0 || cur >= prev)) return false;
                prev = cur;
                if (node.left != null) queue.offer(node.left);
                if (node.right != null) queue.offer(node.right);
            }
            flag = !flag;
        }
        return true;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$

方法三：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 {
    Map<Integer, Integer> map = new HashMap<>();
    
    public boolean isEvenOddTree(TreeNode root) {
        return dfs(root, 0);
    }

    private boolean dfs(TreeNode root, int idx) {
        boolean flag = idx % 2 == 0;
        int prev = map.getOrDefault(idx, flag ? 0 : Integer.MAX_VALUE), cur = root.val;
        if (flag && (cur % 2 == 0 || cur <= prev)) return false;
        if (!flag && (cur % 2 != 0 || cur >= prev)) return false;
        map.put(idx, root.val);
        if (root.left != null && !dfs(root.left, idx + 1)) return false;
        if (root.right != null && !dfs(root.right, idx + 1)) return false;
        return true;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$