LeetCode110：Balanced Binary Tree

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.

---

LeetCode：链接

按照平衡二叉树的定义，有两点需要注意：1，每个节点的两个子树也都是平衡二叉树；2，每个节点的两个子树的高度差不超过1。 直接的思路是：遍历每个节点，判断每个节点的两个子树高度差是否小于等于1（第一个递归）；而求一个子树的高度可以用递归法求解（第二个递归）。

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if root == None:
            return True
        if abs(self.getDepth(root.left) - self.getDepth(root.right)) > 1:
            return False
        return self.isBalanced(root.left) and self.isBalanced(root.right)

    def getDepth(self, root):
        if root == None:
            return 0
        return 1 + max(self.getDepth(root.left), self.getDepth(root.right))

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