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 every node never differ by more than 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 (TreeNode* root) 
    {
        if(root==NULL)
            return true;
        int left=getDepth(root->left);
        int right=getDepth(root->right);
        if(left-right<-1||left-right>1)
        {
            return false;
        }
        return isBalanced(root->left)&&isBalanced(root->right);

    }
    int getDepth(TreeNode* root)
    {
        if(root==NULL)
            return 0;
        int left=getDepth(root->left);
        int right=getDepth(root->right);
        return max(left,right)+1;
    }
};

今天学了点java，练习下。简单看这题的话，java和C++没什么区别，就是把指针换成了引用。

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* Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public boolean (TreeNode root) { if(root==null) return true; int left=getDepth(root.left); int right=getDepth(root.right); if(Math.abs(left-right)>1) { return false; } return isBalanced(root.left)&&isBalanced(root.right); } public int getDepth(TreeNode root) { if(root==null) return 0; int left=getDepth(root.left); int right=getDepth(root.right); return Math.max(left,right)+1; } }