[LeetCode] Balanced Binary Tree 平衡二叉树

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 everynode 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.

 

求二叉树是否平衡，根据题目中的定义，高度平衡二叉树是每一个结点的两个子树的深度差不能超过1，那么我们肯定需要一个求各个点深度的函数，然后对每个节点的两个子树来比较深度差，时间复杂度为O(NlgN)，代码如下：

解法一：

class Solution {
public:
    bool isBalanced(TreeNode *root) {
        if (!root) return true;
        if (abs(getDepth(root->left) - getDepth(root->right)) > ) return false;
        return isBalanced(root->left) && isBalanced(root->right);
    }
    int getDepth(TreeNode *root) {
        if (!root) return ;
        return  + max(getDepth(root->left), getDepth(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checkDepth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：

解法二：

class Solution {
public:
    bool isBalanced(TreeNode *root) {
        if (checkDepth(root) == -) return false;
        else return true;
    }
    int checkDepth(TreeNode *root) {
        if (!root) return ;
        int left = checkDepth(root->left);
        if (left == -) return -;
        int right = checkDepth(root->right);
        if (right == -) return -;
        int diff = abs(left - right);
        if (diff > ) return -;
        else return  + max(left, right);
    }
};

类似题目：

Maximum Depth of Binary Tree

参考资料：

https://leetcode.com/problems/balanced-binary-tree/

https://leetcode.com/problems/balanced-binary-tree/discuss/35691/The-bottom-up-O(N)-solution-would-be-better

https://leetcode.com/problems/balanced-binary-tree/discuss/35686/Java-solution-based-on-height-check-left-and-right-node-in-every-recursion-to-avoid-further-useless-search

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