学步园

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.

code : 

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isBalanced(TreeNode *root) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(root == NULL)
            return true;
        int ld = treedepth(root->left);
        int rd = treedepth(root->right);
        if(abs(ld - rd) > 1)
            return false;
        return isBalanced(root->left) && isBalanced(root->right);
    }
    
    int treedepth(TreeNode *root)
    {
        if(root == NULL) 
            return 0;
        int ldepth = treedepth(root->left);
        int rdepth = treedepth(root->right);
        return ldepth > rdepth ? ldepth+1:rdepth+1; 
    }
};

上面算法效率低，因为重复访问了结点，优化下，改成后序遍历即可。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isBalanced(TreeNode *root) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int depth = 0;
        return isbalance(root, depth);
        
    }
    bool isbalance(TreeNode *root, int &depth)
    {
        if(root == NULL)
        {
            depth = 0;
            return true;
        }
        int ld,rd;
        if( isbalance(root->left,ld) && isbalance(root->right,rd))
        {
            if( abs(ld - rd) > 1)
            {
                return false;
            }
            depth = ld > rd ? ld + 1 : rd + 1;
            return true;
        }
    }
    
    
};