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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。

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。

Python3

# 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.
# 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)

Java

/**
 * 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 : 1000000;
            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.
 * 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 : 1000000);
        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);
    }
}

C++

/**
 * 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 : 1e6;
            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.
 * 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 : 1e6);
        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);
    }
};

Go

/**
 * 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 = 1e6
		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
}
/**
 * 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 = 1000000
			}
		}
		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)
}

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