判断一棵二叉查找树是否是平衡树

来源：互联网发布：cf无后坐力软件编辑：程序博客网时间：2024/05/20 14:23

首先平衡树的定义是，若一棵二叉查找树中任意一个结点的左右子树的深度差不超过1，则该树是平衡的。

法一：在遍历树的过程中，对每个结点的左右子树调用求深度TreeDepth。

结点的结构体：

struct TreeNode;typedef struct TreeNode *Position;typedef struct TreeNode *SearchTree;typedef int ElementType;struct TreeNode{ElementType Data;SearchTree Left;SearchTree Right;};

求树的深度：

int TreeDepth(SearchTree T){if (T == NULL)return 0;else{int Left = TreeDepth(T->Left);int Right = TreeDepth(T->Right);return Left > Right ? (Left + 1) : (Right + 1);}}

判 AVL：

bool isAVLTree_1(SearchTree T){if (T == NULL)return true;else{int left = TreeDepth(T->Left);int right = TreeDepth(T->Right);int diff = left - right;if (diff > 1 || diff < -1)return false;}return isAVLTree_1(T->Left) && isAVLTree_1(T->Right);}

分析：虽然使用上述方法简单易懂，但是一个结点会被重复遍历多次，导致效率低。以下图为例，首先判断根节点是否是平衡时要访问5,6结点，然后在判断3是否是平衡时也要访问5,6。显然，重复遍历一个结点会直接影响性能。

法二：用后序遍历的方式遍历每一个结点

使用后序遍历时，在遍历到一个结点之前已经遍历了它的左右子树，所以只要在遍历每个结点时记录它的深度，就可以一边遍历一边判断每个结点是不是平衡的。

bool isAVLTree_2(SearchTree T, int* depth){if (T == NULL){*depth = 0;return true;}int left, right;if (isAVLTree_2(T->Left, &left) && isAVLTree_2(T->Right, &right)){int diff = left - right;if (diff <= 1 && diff >= -1){*depth = 1 + (left > right ? left : right);return true;}}return false;}bool isAVLTree_2(SearchTree T){int depth = 0;return isAVLTree_2(T, &depth);}

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