判断一颗树是否为平衡二叉树

1. struct BinaryTreeNode
2. {
3. int m_Value;
4. BinaryTreeNode* m_pLeft;
5. BinaryTreeNode* m_pRight;
6. };
7. int height(BinaryTreeNode*node ,bool&balance)
8. {
9. if(node==NULL)
10. {
11. return 0;}
12. int R=node->right?height(node->right,balance)+1:0;
13. if(!balance)
14. return 0;
15. int l=node->left?height(node->left,balance)+1:0;
16. if(!balance)
17. return 0;
18. int dif=l-r;
19. if (dif<=1)
20. {
21. return (l>r?l:r)
22. }
23. }
24. bool  isBalancedTree(node *root)
25. {
26. bool balance=tree;
27. if(root)
28. treehight(root,balance)
29. return balance;
30. }

//递归：

1. struct BinaryTreeNode
2. {
3. int m_Value;
4. BinaryTreeNode* m_pLeft;
5. BinaryTreeNode* m_pRight;
6. };
7. int TreeDepth(BinaryTreeNode* pRoot)
8. {
9. if (pRoot == NULL)
10. return 0;
11. int nLeftDepth = TreeDepth(pRoot->m_pLeft);
12. int nRightDepth = TreeDepth(pRoot->m_pRight);
13. return (nLeftDepth>nRightDepth)?(nLeftDepth+1):(nRightDepth+1);
14. }

判断该树是否为平衡二叉树

方法一：调用上述函数求每个节点的左右孩子深度

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print?

1. bool IsBalanced(BinaryTreeNode* pRoot)
2. {
3. if(pRoot== NULL)
4. return true;
5. int nLeftDepth = TreeDepth(pRoot->m_pLeft);
6. int nRightDepth = TreeDepth(pRoot->m_pRight);
7. int diff = nRightDepth-nLeftDepth;
8. if (diff>1 || diff<-1)
9. return false;
10. return IsBalanced(pRoot->m_pLeft)&&IsBalanced(pRoot->m_pRight);
11. }