110. Balanced Binary Tree

image.png

解法一：由上至下

/**
 * 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:
    int depth(TreeNode* root){
        if(root == NULL){
            return 0;
        }
        int left = depth(root->left);
        int right = depth(root->right);
        return max(left, right) + 1;
    }
    bool isBalanced(TreeNode* root) {
        if(root == NULL){
            return true;
        }
        int left = depth(root->left);
        int right = depth(root->right);
        
        return abs(left - right) <= 1 && isBalanced(root->left) && isBalanced(root->right);
    }
};

解法二：dfs

/**
 * 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:
    int dfsHeight(TreeNode* root){
        if(root == NULL) return 0;
        int left = dfsHeight(root->left);
        if(left == -1) return -1;
        int right = dfsHeight(root->right);
        if(right == -1) return -1;
        if(abs(right - left) > 1) return -1;
        else return max(left, right) + 1;
        }
    bool isBalanced(TreeNode* root) {
        if(root == NULL) return true;
        if(dfsHeight(root) != -1) return true;
        else return false;
    }
};