leetcode 110. Balanced Binary Tree

最新推荐文章于 2023-11-11 23:37:07 发布

转载最新推荐文章于 2023-11-11 23:37:07 发布 · 57 阅读

本文介绍了一种算法，用于判断二叉树是否为高度平衡的树。通过递归方式检查每个节点的左右子树深度差异，确保其不超过1。提供了几种实现方法，包括返回平衡状态和高度的组合，直接标记不平衡情况以及使用-1作为不平衡标志。

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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]:

Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,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))