LeetCode_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 ofevery node never differ by more than 1.

分析：检查每个节点的左右子节点的高度差

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

 优化： 《剑指offer》上又一个优化的算法。使用后序遍历的方式，在遍历的过程中，计算每个节点的深度，并判断是否balanced

/**
 * 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 test(TreeNode *root, int &depth){
        if(NULL == root){
            depth = 0;
            return true;
        }
        int leftD, rightD;
        bool testLeft = test(root->left ,leftD);
        bool testRight = test(root->right, rightD);
        depth  = leftD > rightD ? leftD+1 : rightD +1 ;
        
        if(testLeft && testRight){
            return  abs(leftD - rightD) < 2;
        }        
        return  false;
    }
    bool isBalanced(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int depth;
        return test(root, depth);
    }
};