36.平衡二叉树

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平衡二叉树

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算法知识视频讲解

题目描述

输入一棵二叉树，判断该二叉树是否是平衡二叉树。

这道题承接上一题二叉树的深度，边遍历边使用getDepth()函数。

class Solution {public:    bool IsBalanced_Solution(TreeNode* pRoot) {if ( pRoot == NULL ) return true ;                int left = getDepth( pRoot->left ) ;        int right = getDepth( pRoot->right ) ;                int val = left - right ;        if ( val > 1 || val < -1 ) return false ;        else return IsBalanced_Solution( pRoot->left ) && IsBalanced_Solution( pRoot->right ) ;    }    int getDepth(TreeNode* pNode) {          if (pNode == NULL) return 0;            int depthLeft = getDepth(pNode->left);          int depthRight = getDepth(pNode->right);            return depthLeft > depthRight ? depthLeft + 1 : depthRight + 1;      }  };

第二次做：

跟上一题（树的深度）一样，思路要灵活，递归要精彩~

class Solution {public:    bool IsBalanced_Solution(TreeNode* pRoot) {if ( pRoot == NULL ) return true ;                int left = TreeDepth( pRoot->left ) ;        int right = TreeDepth( pRoot->right ) ;        int diff = left - right ;        if ( diff > 1 || diff < -1 ) return false ;                return IsBalanced_Solution( pRoot->left ) && IsBalanced_Solution( pRoot->right ) ;    }        int TreeDepth( TreeNode* pRoot ) {        if ( pRoot == NULL ) return 0 ;                int left = TreeDepth( pRoot->left ) ;        int right = TreeDepth( pRoot->right ) ;                return left > right ? left + 1 : right + 1 ;    }};

第三次做：

class Solution {public:    bool IsBalanced_Solution(TreeNode* pRoot) {if ( pRoot == NULL ) return true ;                int left = tree_depth( pRoot->left ) ;        int right = tree_depth( pRoot->right ) ;                int step = left - right ;        if ( step > 1 || step < -1 ) return false ;                return IsBalanced_Solution( pRoot->left ) && IsBalanced_Solution( pRoot->right ) ;    }        int tree_depth( TreeNode* root ) {        if ( root == NULL ) return 0 ;                int left = tree_depth( root->left ) ;        int right = tree_depth( root->right ) ;                return left > right ? left + 1 : right + 1 ;    }};

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