leetcode 110 Balanced Binary Tree

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.

 

树的高度： Tree depth的定义：为当前节点左右subtree depth中的较大值加1.

// 判断左右子树是不是平衡，然后 左右子树的高度差不大于1

// 可以把树深度的返回 和 判断平衡树 放在一起
// 用一个特殊的返回值来代表不平衡，这样就少了一层的判断

// Tree depth的定义：为当前节点左右subtree depth中的较大值加1.

从根往上

class Solution {
public:
    bool isBalanced(TreeNode* root) {
        return dfs(root) == -1? false: true;
    }
    int dfs(TreeNode* root)
    {
        if (root == NULL)
            return 0;
        int left = dfs(root -> left);
        int right = dfs(root -> right);
        if (left == -1 || right == -1|| abs(left - right) >1) return -1;
        return max(left,right) +1;
        
    }
};

 

方法二： 

从上往下，但是会消耗跟多的时间

class Solution {
public:
    bool isBalanced(TreeNode* root) {
        if (root == NULL)
            return true;
        if (isBalanced(root->left) && isBalanced(root->right))
            return abs(dfs(root->left) - dfs(root->right)) >1 ? false:true; 
        return false; 
    }
    int dfs(TreeNode* root)
    {
        if (root == NULL)
            return 0;
        if (root ->left == NULL || root->right == NULL )
            return dfs(root->left) + dfs(root ->right) +1 ;
        return max(dfs(root->left),dfs(root->right))+1;
    }
};