标签：ble type block amp max als not bin 高度
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as:
a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example 1:
Given the following tree [3,9,20,null,null,15,7]:
3
/ 9 20
/ 15 7
Return true.
Example 2:
Given the following tree [1,2,2,3,3,null,null,4,4]:
1
/ 2 2
/ 3 3
/ 4 4
Return false.
# 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):
"""
:type root: TreeNode
:rtype: bool
"""
def helper(node):
if not node:
return True, 0
l_b, l_h = helper(node.left)
r_b, r_h = helper(node.right)
if not l_b or not r_b or abs(l_h-r_h) > 1:
return False, max(l_h, r_h)+1
return True, max(l_h, r_h)+1
ans, _ = helper(root)
return ans
或者是直接在获取tree高度的时候用一个成员变量来记录是否平衡。
# 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):
"""
:type root: TreeNode
:rtype: bool
"""
self.ans = True
def get_depth(node):
if not node:
return 0
l, r = get_depth(node.left), get_depth(node.right)
if abs(l-r) > 1:
self.ans = False
return max(l, r)+1
get_depth(root)
return self.ans
另外的解法就是使用高度是否为-1来标识是否平衡。
class solution {
public:
int dfsHeight (TreeNode *root) {
if (root == NULL) return 0;
int leftHeight = dfsHeight (root -> left);
if (leftHeight == -1) return -1;
int rightHeight = dfsHeight (root -> right);
if (rightHeight == -1) return -1;
if (abs(leftHeight - rightHeight) > 1) return -1;
return max (leftHeight, rightHeight) + 1;
}
bool isBalanced(TreeNode *root) {
return dfsHeight (root) != -1;
}
};
最笨的方法就是N^2复杂度的：
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# @param {TreeNode} root
# @return {boolean}
def isBalanced(self, root):
if not root:
return True
return abs(self.getHeight(root.left) - self.getHeight(root.right)) < 2 and self.isBalanced(root.left) and self.isBalanced(root.right)
def getHeight(self, root):
if not root:
return 0
return 1 + max(self.getHeight(root.left), self.getHeight(root.right))
leetcode 110. Balanced Binary Tree
标签：ble type block amp max als not bin 高度
原文地址：https://www.cnblogs.com/bonelee/p/9180555.html
