平衡二叉树

问题

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.

思路

递归计算每一个节点的深度。简单粗暴，但存在的问题是，每一个点都会被上面的点计算深度时重新递归计算一次。
判断是否平衡，采用前序遍历。若当前节点平衡，不返回值，继续向下遍历判断。直到节点为空，不断向上返回true。

class Solution {
public:
    int nodeDepth(TreeNode* root)
    {
        if(root == NULL)
        {
            return 0;
        }
        return 1 + max(nodeDepth(root->left), nodeDepth(root->right));
    }

    bool isBalanced(TreeNode* root) {
        if(root == NULL)
        {
            return true;
        }
        if(abs(nodeDepth(root->left) - nodeDepth(root->right)) >= 2)
        {
            return false;
        }
        return isBalanced(root->left) && isBalanced(root->right);
    }
};

优化上面的递归，发现子树不平衡，则不再计算深度，而是直接返回-1

class Solution {
public:    
    bool isBalanced(TreeNode *root) {
        if (checkDepth(root) == -1) return false;
        else return true;
    }
    int checkDepth(TreeNode *root) {
        if (!root) return 0;
        int left = checkDepth(root->left);
        if (left == -1) return -1;
        int right = checkDepth(root->right);
        if (right == -1) return -1;
        int diff = abs(left - right);
        if (diff > 1) return -1;
        else return 1 + max(left, right);
    }
};