Question

1609. 奇偶树

English Version

题目描述

如果一棵二叉树满足下述几个条件，则可以称为 奇偶树 ：

• 二叉树根节点所在层下标为 0 ，根的子节点所在层下标为 1 ，根的孙节点所在层下标为 2 ，依此类推。
• 偶数下标 层上的所有节点的值都是 奇 整数，从左到右按顺序 严格递增
• 奇数下标 层上的所有节点的值都是 偶 整数，从左到右按顺序 严格递减

给你二叉树的根节点，如果二叉树为 奇偶树 ，则返回 true ，否则返回 false 。

 

示例 1：

输入：root = [1,10,4,3,null,7,9,12,8,6,null,null,2]
输出：true
解释：每一层的节点值分别是：
0 层：[1]
1 层：[10,4]
2 层：[3,7,9]
3 层：[12,8,6,2]
由于 0 层和 2 层上的节点值都是奇数且严格递增，而 1 层和 3 层上的节点值都是偶数且严格递减，因此这是一棵奇偶树。

示例 2：

输入：root = [5,4,2,3,3,7]
输出：false
解释：每一层的节点值分别是：
0 层：[5]
1 层：[4,2]
2 层：[3,3,7]
2 层上的节点值不满足严格递增的条件，所以这不是一棵奇偶树。

示例 3：

输入：root = [5,9,1,3,5,7]
输出：false
解释：1 层上的节点值应为偶数。

示例 4：

输入：root = [1]
输出：true

示例 5：

输入：root = [11,8,6,1,3,9,11,30,20,18,16,12,10,4,2,17]
输出：true

 

提示：

• 树中节点数在范围 [1, 105] 内
• 1 <= Node.val <= 106

解法

方法一：BFS

BFS 逐层遍历，每层按照奇偶性判断，每层的节点值都是偶数或奇数，且严格递增或递减。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。

# 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 isEvenOddTree(self, root: Optional[TreeNode]) -> bool:
    even = 1
    q = deque([root])
    while q:
      prev = 0 if even else inf
      for _ in range(len(q)):
        root = q.popleft()
        if even and (root.val % 2 == 0 or prev >= root.val):
          return False
        if not even and (root.val % 2 == 1 or prev <= root.val):
          return False
        prev = root.val
        if root.left:
          q.append(root.left)
        if root.right:
          q.append(root.right)
      even ^= 1
    return True

/**
 * 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) {
    boolean even = true;
    Deque<TreeNode> q = new ArrayDeque<>();
    q.offer(root);
    while (!q.isEmpty()) {
      int prev = even ? 0 : 1000001;
      for (int n = q.size(); n > 0; --n) {
        root = q.pollFirst();
        if (even && (root.val % 2 == 0 || prev >= root.val)) {
          return false;
        }
        if (!even && (root.val % 2 == 1 || prev <= root.val)) {
          return false;
        }
        prev = root.val;
        if (root.left != null) {
          q.offer(root.left);
        }
        if (root.right != null) {
          q.offer(root.right);
        }
      }
      even = !even;
    }
    return true;
  }
}

/**
 * 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:
  bool isEvenOddTree(TreeNode* root) {
    int even = 1;
    queue<TreeNode*> q{{root}};
    while (!q.empty()) {
      int prev = even ? 0 : 1e7;
      for (int n = q.size(); n; --n) {
        root = q.front();
        q.pop();
        if (even && (root->val % 2 == 0 || prev >= root->val)) {
          return false;
        }
        if (!even && (root->val % 2 == 1 || prev <= root->val)) {
          return false;
        }
        prev = root->val;
        if (root->left) {
          q.push(root->left);
        }
        if (root->right) {
          q.push(root->right);
        }
      }
      even ^= 1;
    }
    return true;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func isEvenOddTree(root *TreeNode) bool {
  even := true
  q := []*TreeNode{root}
  for len(q) > 0 {
    var prev int = 1e7
    if even {
      prev = 0
    }
    for n := len(q); n > 0; n-- {
      root = q[0]
      q = q[1:]
      if even && (root.Val%2 == 0 || prev >= root.Val) {
        return false
      }
      if !even && (root.Val%2 == 1 || prev <= root.Val) {
        return false
      }
      prev = root.Val
      if root.Left != nil {
        q = append(q, root.Left)
      }
      if root.Right != nil {
        q = append(q, root.Right)
      }
    }
    even = !even
  }
  return true
}

方法二：DFS

DFS 先序遍历二叉树，同样根据节点所在层的奇偶性判断是否满足条件，遍历过程中用哈希表记录每一层最近访问到的节点值。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉树的节点数。

# 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 isEvenOddTree(self, root: Optional[TreeNode]) -> bool:
    def dfs(root, i):
      if root is None:
        return True
      even = i % 2 == 0
      prev = d.get(i, 0 if even else inf)
      if even and (root.val % 2 == 0 or prev >= root.val):
        return False
      if not even and (root.val % 2 == 1 or prev <= root.val):
        return False
      d[i] = root.val
      return dfs(root.left, i + 1) and dfs(root.right, i + 1)

    d = {}
    return dfs(root, 0)

/**
 * 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 Map<Integer, Integer> d = new HashMap<>();

  public boolean isEvenOddTree(TreeNode root) {
    return dfs(root, 0);
  }

  private boolean dfs(TreeNode root, int i) {
    if (root == null) {
      return true;
    }
    boolean even = i % 2 == 0;
    int prev = d.getOrDefault(i, even ? 0 : 1000001);
    if (even && (root.val % 2 == 0 || prev >= root.val)) {
      return false;
    }
    if (!even && (root.val % 2 == 1 || prev <= root.val)) {
      return false;
    }
    d.put(i, root.val);
    return dfs(root.left, i + 1) && dfs(root.right, i + 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:
  unordered_map<int, int> d;

  bool isEvenOddTree(TreeNode* root) {
    return dfs(root, 0);
  }

  bool dfs(TreeNode* root, int i) {
    if (!root) {
      return true;
    }
    int even = i % 2 == 0;
    int prev = d.count(i) ? d[i] : (even ? 0 : 1e7);
    if (even && (root->val % 2 == 0 || prev >= root->val)) {
      return false;
    }
    if (!even && (root->val % 2 == 1 || prev <= root->val)) {
      return false;
    }
    d[i] = root->val;
    return dfs(root->left, i + 1) && dfs(root->right, i + 1);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func isEvenOddTree(root *TreeNode) bool {
  d := map[int]int{}
  var dfs func(*TreeNode, int) bool
  dfs = func(root *TreeNode, i int) bool {
    if root == nil {
      return true
    }
    even := i%2 == 0
    prev, ok := d[i]
    if !ok {
      if even {
        prev = 0
      } else {
        prev = 10000000
      }
    }
    if even && (root.Val%2 == 0 || prev >= root.Val) {
      return false
    }
    if !even && (root.Val%2 == 1 || prev <= root.Val) {
      return false
    }
    d[i] = root.Val
    return dfs(root.Left, i+1) && dfs(root.Right, i+1)
  }
  return dfs(root, 0)
}