letecode [110] - Balanced Binary Tree

发布时间：2020-12-14 04:44:44 所属栏目：大数据来源：网络整理

导读：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. Example 1: Given the foll

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.

Example 1:

Given the following tree [3,9,20,null,15,7]:

??? 3
?? /
? 9? 20
??? /?
?? 15?? 7

Return true.

Example 2:

Given the following tree [1,2,3,4,4]:

?????? 1
????? /
???? 2?? 2
??? /
?? 3?? 3
? /
?4?? 4

Return false.

题目大意：

　　给定一个二叉树，判断该二叉树是否为平衡二叉树。

理? 解：

　　判断二叉树是否为平衡二叉树返回的是bool值。而在判断过程中需要比较当前节点的左右子树高度之差是否超过1.

　　用另一个函数，求得当前左右子树的高度，若相差超过1，则返回INT_MAX，即不是平衡二叉树。

代码 C++：

/**
 * 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 m_isBalanced(TreeNode* root){
        
        if(root==NULL) return 0;
        if(root->left==NULL && right==NULL) return 1;
        int left,right;
        left = m_isBalanced(root->left);
        right = m_isBalanced(root->right);
        if(left==INT_MAX || right==INT_MAX)
            return INT_MAX;
        if(left-right>1 || left-right<-1) 
            return INT_MAX;
        return left>right? left+1:right+1;
    }
    
    bool isBalanced(TreeNode* root) {
        if(root==NULL) return true;
        if(root->left==NULL && right==NULL) return true;
        if(m_isBalanced(root)==INT_MAX)   
            return false;
        return true;
    }
};

运行结果：

　　执行用时 :?8 ms　　内存消耗 :?16.7 MB

（编辑：李大同）

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