letecode [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.

Example 1:

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

    3
   / \
  9  20
    /  \
   15   7

Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,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