[LeetCode]110. Balanced Binary Tree

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Problem

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]:

1
2
3
4
5
  3
 / \
9  20
  /  \
 15   7

Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

1
2
3
4
5
6
7
      1
     / \
    2   2
   / \
  3   3
 / \
4   4

Return false.

Solution

解析

根据平衡二叉树定义，每个节点子树之间的高度差不能超过1.所以我们先定义一个求二叉树高度的函数getDepth()，然后递归对二叉树节点进行判断。

Solution 1

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isBalanced(TreeNode* root) {
        if (!root) return true;
        int left = getDepth(root->left), right = getDepth(root->right);
        
        if ( left - right > 1 || right - left > 1) return false;
        
        return isBalanced(root->left) && isBalanced(root->right);
    }
private:
    int getDepth(TreeNode* root){
        if (!root) return 0;
        int left = getDepth(root->left);
        int right = getDepth(root->right);
        
        return max(left, right) + 1;
    }
};

这种方法速度比较慢。原因在于，节点高度的重复计算。自上而下，节点的高度多次重复计算，提高了时间复杂度。

如果简化的话，想办法自下而上，减少重复计算。这里，修改getDepth()函数，让函数在求二叉树深度的同时，判断左右节点的高度差，如果大于1，返回-1(说明这棵二叉树一定不是平衡二叉树，及时停止)。

Solution 2

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isBalanced(TreeNode* root) {
        return dfsDepth(root) != -1;
    }
private:
    int dfsDepth(TreeNode* root){
        if (!root) return 0;
        int left = dfsDepth(root->left);//递归函数
        int right = dfsDepth(root->right);
        
        if (left == -1 || right == -1) return -1;//子树不是avl
        if (abs(left - right) > 1) return -1;//当前树不是avl
        
        return max(left, right) + 1;
    }
};
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