平衡的二叉树的定义都是递归的定义,所以,用递归来解决问题,还是挺容易的额。
本质上是递归的遍历二叉树。
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给定一个二叉树,判定他是不是高度平衡的二叉树。
对于这个问题,每个节点的两个子树的深度不会相差超过1,那么这样的二叉树就是一个平衡的二叉树
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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.
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test.cpp:
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#include <iostream> #include <cstdio> #include <stack> #include <vector> #include "BinaryTree.h" using namespace std; /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ bool balance(TreeNode *root, int &depth) { if(root == NULL) { depth = 0; return true; } int leftdepth = 0; bool left = balance(root->left, leftdepth); int rightdepth = 0; bool right = balance(root->right, rightdepth); depth = leftdepth > rightdepth ? leftdepth + 1 : rightdepth + 1; int gap = leftdepth - rightdepth; return left && right && (-1 <= gap && gap <= 1); } bool isBalanced(TreeNode *root) { int depth = 0; return balance(root, depth); } // 树中结点含有分叉, // 6 // / \ // 7 2 // / \ // 1 4 // / \ // 3 5 int main() { TreeNode *pNodeA1 = CreateBinaryTreeNode(6); TreeNode *pNodeA2 = CreateBinaryTreeNode(7); TreeNode *pNodeA3 = CreateBinaryTreeNode(2); TreeNode *pNodeA4 = CreateBinaryTreeNode(1); TreeNode *pNodeA5 = CreateBinaryTreeNode(4); TreeNode *pNodeA6 = CreateBinaryTreeNode(3); TreeNode *pNodeA7 = CreateBinaryTreeNode(5); ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3); ConnectTreeNodes(pNodeA2, pNodeA4, pNodeA5); ConnectTreeNodes(pNodeA5, pNodeA6, pNodeA7); // 树中结点含有分叉, // 1 // / \ // 2 2 // / \ / \ // 3 4 4 3 TreeNode *pNodeB1 = CreateBinaryTreeNode(1); TreeNode *pNodeB2 = CreateBinaryTreeNode(2); TreeNode *pNodeB3 = CreateBinaryTreeNode(2); TreeNode *pNodeB4 = CreateBinaryTreeNode(3); TreeNode *pNodeB5 = CreateBinaryTreeNode(4); TreeNode *pNodeB6 = CreateBinaryTreeNode(4); TreeNode *pNodeB7 = CreateBinaryTreeNode(3); ConnectTreeNodes(pNodeB1, pNodeB2, pNodeB3); ConnectTreeNodes(pNodeB2, pNodeB4, pNodeB5); ConnectTreeNodes(pNodeB3, pNodeB6, pNodeB7); bool ans = isBalanced(pNodeA1); if (ans == true) { cout << "Balanced!" << endl; } else { cout << "Not Balanced!" << endl; } bool ans1 = isBalanced(pNodeB1); if (ans1 == true) { cout << "Balanced!" << endl; } else { cout << "Not Balanced!" << endl; } DestroyTree(pNodeA1); DestroyTree(pNodeB1); return 0; } |
结果输出:
Not Balanced!
Balanced!
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#ifndef _BINARY_TREE_H_ #define _BINARY_TREE_H_ struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode(int x) : val(x), left(NULL), right(NULL) {} }; TreeNode *CreateBinaryTreeNode(int value); void ConnectTreeNodes(TreeNode *pParent, TreeNode *pLeft, TreeNode *pRight); void PrintTreeNode(TreeNode *pNode); void PrintTree(TreeNode *pRoot); void DestroyTree(TreeNode *pRoot); #endif /*_BINARY_TREE_H_*/ |
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#include <iostream> #include <cstdio> #include "BinaryTree.h" using namespace std; /** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ //创建结点 TreeNode *CreateBinaryTreeNode(int value) { TreeNode *pNode = new TreeNode(value); return pNode; } //连接结点 void ConnectTreeNodes(TreeNode *pParent, TreeNode *pLeft, TreeNode *pRight) { if(pParent != NULL) { pParent->left = pLeft; pParent->right = pRight; } } //打印节点内容以及左右子结点内容 void PrintTreeNode(TreeNode *pNode) { if(pNode != NULL) { printf("value of this node is: %d\n", pNode->val); if(pNode->left != NULL) printf("value of its left child is: %d.\n", pNode->left->val); else printf("left child is null.\n"); if(pNode->right != NULL) printf("value of its right child is: %d.\n", pNode->right->val); else printf("right child is null.\n"); } else { printf("this node is null.\n"); } printf("\n"); } //前序遍历递归方法打印结点内容 void PrintTree(TreeNode *pRoot) { PrintTreeNode(pRoot); if(pRoot != NULL) { if(pRoot->left != NULL) PrintTree(pRoot->left); if(pRoot->right != NULL) PrintTree(pRoot->right); } } void DestroyTree(TreeNode *pRoot) { if(pRoot != NULL) { TreeNode *pLeft = pRoot->left; TreeNode *pRight = pRoot->right; delete pRoot; pRoot = NULL; DestroyTree(pLeft); DestroyTree(pRight); } } |
【遍历二叉树】10判断二叉树是否平衡【Balanced Binary Tree】,布布扣,bubuko.com
【遍历二叉树】10判断二叉树是否平衡【Balanced Binary Tree】
原文:http://www.cnblogs.com/codemylife/p/3652344.html