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 ofeverynode never differ by more than 1.

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

Given binary tree A={3,9,20,#,#,15,7}, B={3,#,20,15,7}

A)  3            B)    3
   / \                  \
  9  20                 20
    /  \                / \
   15   7              15  7

The binary tree A is a height-balanced binary tree, but B is not.

这一题也是用recursive的方法做

• 对于每一个root,分别get左右两边子树的max depth

• 如果balanced，就继续recursive的检查左右子树

• 时间复杂度为O(NlgN)

# 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 not root:
            return True
        if abs(self.depth(root.left)-self.depth(root.right)) > 1:
            return False
        return self.isBalanced(root.left) and self.isBalanced(root.right)

    def depth(self, root):
        if not root:
            return 0
        return 1+max(self.depth(root.left),self.depth(root.right))

上面那个方法正确但不是很高效，因为每一个点都会被上面的点计算深度时访问一次，我们可以进行优化。方法是如果我们发现子树不平衡，则不计算具体的深度，而是直接返回-1。那么优化后的方法为：对于每一个节点，我们通过check方法递归获得左右子树的深度，如果子树是平衡的，则返回真实的深度，若不平衡，直接返回-1，此方法时间复杂度O(N)，空间复杂度O(H)，参见代码如下：

https://www.youtube.com/watch?v=TWDigbwxuB4

# 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
        """
        def check(root):
            if not root:
                return 0
            leftnode = check(root.left)
            if leftnode == -1:
                return -1
            rightnode = check(root.right)
            if rightnode == -1:
                return -1
            if abs(leftnode - rightnode)>1:
                return -1
            return 1+max(leftnode,rightnode)
        return check(root) != -1