1609. Even Odd Tree

Description

A binary tree is named Even-Odd if it meets the following conditions:

The root of the binary tree is at level index 0, its children are at level index 1, their children are at level index 2, etc.
For every even-indexed level, all nodes at the level have odd integer values in strictly increasing order (from left to right).
For every odd-indexed level, all nodes at the level have even integer values in strictly decreasing order (from left to right).

Given the root of a binary tree, return true if the binary tree is Even-Odd, otherwise return false.

Example 1:

Input: root = [1,10,4,3,null,7,9,12,8,6,null,null,2]
Output: true
Explanation: The node values on each level are:
Level 0: [1]
Level 1: [10,4]
Level 2: [3,7,9]
Level 3: [12,8,6,2]
Since levels 0 and 2 are all odd and increasing and levels 1 and 3 are all even and decreasing, the tree is Even-Odd.

Example 2:

Input: root = [5,4,2,3,3,7]
Output: false
Explanation: The node values on each level are:
Level 0: [5]
Level 1: [4,2]
Level 2: [3,3,7]
Node values in level 2 must be in strictly increasing order, so the tree is not Even-Odd.

Example 3:

Input: root = [5,9,1,3,5,7]
Output: false
Explanation: Node values in the level 1 should be even integers.

Constraints:

The number of nodes in the tree is in the range [1, 10⁵].
1 <= Node.val <= 10⁶

Solutions

BFS.

/**
 * 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) {
    Queue < TreeNode > que = new LinkedList < > ();
    que.add(root);
    que.add(null);
    int res = 1;
    int c = 0;
    int prev = 0;
    while (!que.isEmpty()) {
      if (res % 2 == 0)
        prev = Integer.MAX_VALUE;
      else
        prev = Integer.MIN_VALUE;
      while (que.peek() != null) {
        if (que.peek().val % 2 != res % 2) {
          return false;
        }
        if (res % 2 == 0 && prev <= que.peek().val) {
          return false;
        }

        if (res % 2 != 0 && prev >= que.peek().val)
          return false;
        if (que.peek().left != null)
          que.add(que.peek().left);
        if (que.peek().right != null)
          que.add(que.peek().right);
        prev = que.poll().val;
      }
      que.poll();
      if (que.isEmpty()) {
        break;
      }
      res++;
      que.add(null);

    }
    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 : 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.
# 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.
 * 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.
 * 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.
# 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.
 * 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)
}

1609 - Even Odd Tree

1609. Even Odd Tree

Description

Solutions

All Problems

All Solutions

Welcome to Subscribe On Youtube

1609. Even Odd Tree

Description

Solutions

All Problems

All Solutions