题目
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
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/even-odd-tree
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
题解
class Solution {
int[] tmp = new int[100000];
public boolean isEvenOddTree(TreeNode root) {
if(root == null) return false;
return traverseTree(root, 0);
}
public boolean traverseTree(TreeNode root, int level){
if(root == null){
return true;
}else if((level%2 == 0 && (root.val %2 == 0 || (tmp[level] != 0 && root.val <= tmp[level])))
|| (level%2 ==1 && (root.val%2 ==1 || (tmp[level] !=0 && root.val >= tmp[level])))){
return false;
}
tmp[level] = root.val;
boolean l = traverseTree(root.left, level+1);
if(!l) return false;
return traverseTree(root.right, level+1);
}
}执行结果：通过
执行用时：8 ms, 在所有 Java 提交中击败了98.70% 的用户
内存消耗：62.2 MB, 在所有 Java 提交中击败了20.43% 的用户这里用了递归对树进行前序遍历，使用数组保存每层上次遍历到的节点值，内存消耗较多
官方题解
public boolean isEvenOddTree(TreeNode root) {
Queue<TreeNode> queue = new ArrayDeque<TreeNode>();
queue.offer(root);
int level = 0;
while (!queue.isEmpty()) {
int size = queue.size();
int prev = level % 2 == 0 ? Integer.MIN_VALUE : Integer.MAX_VALUE;
for (int i = 0; i < size; i++) {
TreeNode node = queue.poll();
int value = node.val;
if (level % 2 == value % 2) {
return false;
}
if ((level % 2 == 0 && value <= prev) || (level % 2 == 1 && value >= prev)) {
return false;
}
prev = value;
if (node.left != null) {
queue.offer(node.left);
}
if (node.right != null) {
queue.offer(node.right);
}
}
level++;
}
return true;
}使用双端数组队列和迭代方式。
执行用时：9 ms, 在所有 Java 提交中击败了96.96% 的用户
内存消耗：54.4 MB, 在所有 Java 提交中击败了80.00% 的用户速度差不多，内存消耗少了很多。