【遍历二叉树】10判断二叉树是否平衡【Balanced Binary Tree】
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; /** int rightdepth = 0; depth = leftdepth > rightdepth ? leftdepth + 1 : rightdepth + 1; return left && right && (-1 <= gap && gap <= 1); bool isBalanced(TreeNode *root) int depth = 0; // 树中结点含有分叉, ConnectTreeNodes(pNodeA1, pNodeA2, pNodeA3); // 树中结点含有分叉, TreeNode *pNodeB1 = CreateBinaryTreeNode(1); ConnectTreeNodes(pNodeB1, pNodeB2, pNodeB3); bool ans = isBalanced(pNodeA1); if (ans == true) bool ans1 = isBalanced(pNodeB1); if (ans1 == true) |
结果输出:
Not Balanced!
Balanced!
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#ifndef _BINARY_TREE_H_
#define _BINARY_TREE_H_ struct TreeNode TreeNode *CreateBinaryTreeNode(int value); #endif /*_BINARY_TREE_H_*/ |
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#include <iostream>
#include <cstdio> #include "BinaryTree.h" using namespace std; /** //创建结点 return pNode; //连接结点 //打印节点内容以及左右子结点内容 if(pNode->left != NULL) if(pNode->right != NULL) printf("\n"); //前序遍历递归方法打印结点内容 if(pRoot != NULL) if(pRoot->right != NULL) void DestroyTree(TreeNode *pRoot) delete pRoot; DestroyTree(pLeft); |
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