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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.
判断一棵树是否为平衡二叉树。
思路:典型的递归,从第一个根节点开始,看左子树和右子树的深度差值是否小于等于1,然后再看由根节点的左孩子和右孩子构成的两课树是否都符合条件。
代码:
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { private: int maxdepth; public: void solve(TreeNode *root,int depth){ if(root==NULL) return; if(depth>maxdepth) maxdepth=depth; solve(root->left,depth+1); solve(root->right,depth+1); } int getDepth(TreeNode *root) { maxdepth=0; solve(root,1); return maxdepth; } bool isBalanced(TreeNode *root) { if(root==NULL) return true; int l,r; if(root->left!=NULL){ l=getDepth(root->left); }else{ l=0; } if(root->right!=NULL){ r=getDepth(root->right); }else{ r=0; } return abs(l-r)<=1&&isBalanced(root->left)&&isBalanced(root->right); } };
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原文地址:http://www.cnblogs.com/fightformylife/p/4079496.html