LintCode 93: Balanced Binary Tree (Leetcode-110)

最新推荐文章于 2021-01-15 10:18:15 发布

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本文介绍了两种不同的方法来判断二叉树是否为平衡二叉树。第一种方法使用递归来计算每个节点的深度，第二种方法则通过自定义的数据结构ResultType结合深度和平衡性标志进行判断。

解法1：递归。

#include <iostream>
#include <algorithm>

using namespace std;
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

int depth(TreeNode* node) {
    if (!node) return 0;
    return max(depth(node->left), depth(node->right)) + 1;
}


bool isBalanced(TreeNode* root) {
    if (!root) return true;
    if (isBalanced(root->left) && isBalanced(root->right) && (abs(depth(root->left)-depth(root->right))<=1)) return true;
    else return false;
}

int main()
{
    TreeNode t1(3);
    TreeNode t2(9);
    TreeNode t3(20);
    TreeNode t4(15);
    TreeNode t5(7);

    t1.left=&t2;
    t1.right=&t3;
    t3.left=&t4;
    t3.right=&t5;

    cout<<isBalanced(&t1)<<endl;

    TreeNode ta(1);
    TreeNode tb(2);
    TreeNode tc(2);
    TreeNode td(3);
    TreeNode te(3);
    TreeNode tf(4);
    TreeNode tg(4);

    ta.left=&tb;
    ta.right=&tc;
    tb.left=&td;
    tb.right=&te;
    td.left=&tf;
    td.right=&tg;
    cout<<isBalanced(&ta)<<endl;
    return 0;
}

重写了一下，看起来清楚点：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: The root of binary tree.
     * @return: True if this Binary tree is Balanced, or false.
     */
    bool isBalanced(TreeNode * root) {
        if (!root) return true;
        if (!root->left && !root->right) return true;
        if (abs(depth(root->left) - depth(root->right)) > 1) return false;
        return isBalanced(root->left) && isBalanced(root->right);
    }
    
private:
    int depth(TreeNode * root) {
        if (!root) return 0;
        if (!root->left && !root->right) return 1;
        return max(depth(root->left), depth(root->right)) + 1;
    }
};

解法2：用ResultType将bool和depth绑定到一起。
代码如下：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
 
struct ResultType {
    int depth;
    bool isBalanced;
    ResultType(int dp = 0, bool iB = true) : depth(dp), isBalanced(iB) {}
};

class Solution {
public:
    /**
     * @param root: The root of binary tree.
     * @return: True if this Binary tree is Balanced, or false.
     */
    bool isBalanced(TreeNode * root) {
        return depth(root).isBalanced;
    }
    
private:
    ResultType depth(TreeNode * root) {
        if (!root) return ResultType();
        if (!root->left && !root->right) return ResultType(true, 1);
        ResultType left = depth(root->left);
        ResultType right = depth(root->right);
        
        int maxDepth = max(left.depth, right.depth);
        if (abs(left.depth - right.depth) <= 1 && left.isBalanced && right.isBalanced) {
            return ResultType(maxDepth + 1, true);
        } else {
            return ResultType(maxDepth + 1, false);
        }
    }
};

下面这个写法更快。

class Solution {
public:
    /**
     * @param root: The root of binary tree.
     * @return: True if this Binary tree is Balanced, or false.
     */
    bool isBalanced(TreeNode * root) {
        return depth(root).isBalanced;
    }
    
private:
    ResultType depth(TreeNode * root) {
        if (!root) return ResultType();
        
        ResultType left = depth(root->left);
        if (!left.isBalanced) return ResultType(-1, false);
        
        ResultType right = depth(root->right);
        if (!right.isBalanced) return ResultType(-1, false);
        
        int maxDepth = max(left.depth, right.depth);
        if (abs(left.depth - right.depth) <= 1) {
            return ResultType(maxDepth + 1, true);
        } else {
            return ResultType(maxDepth + 1, false);
        }
    }
};

注意：下面这种写法不对，因为left和right没有判断是否isBalanced。

    ResultType depth(TreeNode * root) {
        if (!root) return ResultType();
        if (!root->left && !root->right) return ResultType(true, 1);
        ResultType left = depth(root->left);
        ResultType right = depth(root->right);
        
        int maxDepth = max(left.depth, right.depth);
        if (abs(left.depth - right.depth) > 1) {
            return ResultType(maxDepth + 1, false);
        } else {
            return ResultType(maxDepth + 1, true);
        }
    }