LeetCode | Balanced Binary Tree（平衡二叉树）

2024年1月12日 232次阅读

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.

题目解析：

判断是否是平衡二叉树，也是利用递归的方法，当左右子树都是平衡二叉树，且左右子树高度差小于2的时候，为true，其他都为false。

class Solution {
public:
    bool isBalanced(TreeNode *root) {
        int height;

        return JudgeBalance(root,height);
    }
    bool JudgeBalance(TreeNode *root,int &height){
        if(root == NULL){
            height = 0;
            return true;
        }
        int lheight,rheight;
        bool Judge = JudgeBalance(root->left,lheight) && JudgeBalance(root->right,rheight);
        if(Judge == true){  //写if语句要写全，else也要带上，不然产生不易发现的错误
            height = lheight>rheight ? (lheight+1) : (rheight+1);
            if(lheight-rheight >= 2 || rheight-lheight>=2)
                return false;
        }else
            return false;

        return true;
    }
};

另外的解法：

将bool变量设成私有变量，然后再函数的入口判断，如果为false直接退出，就不递归了。代码很简洁，应学习：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    bool balanced = true;
    
    int height(TreeNode *root) {
        if (!balanced) {
            return -1;
        }        
        if (root == NULL) {
            return 0;
        }
        int lH = height(root->left) + 1;
        int lL = height(root->right) + 1;
        if (abs(lH - lL) > 1) {
            balanced = false;
        }
        return lH > lL ? lH : lL;
    }
    
public:
    bool isBalanced(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        balanced = true;
        height(root);
        return balanced;
    }
};


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