输入一棵二叉树的根节点,判断该树是不是平衡二叉树。如果某二叉树中任意节点的左右子树的深度相差不超过 1,那么它就是一棵平衡二叉树。
示例 1:
给定二叉树 [3,9,20,null,null,15,7]
3
/ \
9 20
/ \
15 7
返回 true
。
示例 2:
给定二叉树 [1,2,2,3,3,null,null,4,4]
1
/ \
2 2
/ \
3 3
/ \
4 4
返回 false
。
限制:
1 <= 树的结点个数 <= 10000
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def isBalanced(self, root: TreeNode) -> bool:
if root is None:
return True
return abs(self._height(root.left) - self._height(root.right)) <= 1 and self.isBalanced(root.left) and self.isBalanced(root.right)
def _height(self, tree):
if tree is None:
return 0
return 1 + max(self._height(tree.left), self._height(tree.right))
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isBalanced(TreeNode root) {
if (root == null) {
return true;
}
return Math.abs(height(root.left) - height(root.right)) <= 1 && isBalanced(root.left) && isBalanced(root.right);
}
private int height(TreeNode tree) {
if (tree == null) {
return 0;
}
return 1 + Math.max(height(tree.left), height(tree.right));
}
}