letecode [110] - Balanced Binary Tree
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.
Example 1:
Given the following tree [3,9,20,null,null,15,7]:
3
/ \
9 20
/ \
15 7
/ \
9 20
/ \
15 7
Return true.
Example 2:
Given the following tree [1,2,2,3,3,null,null,4,4]:
1
/ \
2 2
/ \
3 3
/ \
4 4
/ \
2 2
/ \
3 3
/ \
4 4
Return false.
题目大意:
给定一个二叉树,判断该二叉树是否为平衡二叉树。
理 解 :
判断二叉树是否为平衡二叉树返回的是bool值。而在判断过程中需要比较当前节点的左右子树高度之差是否超过1.
用另一个函数,求得当前左右子树的高度,若相差超过1,则返回INT_MAX,即不是平衡二叉树。
代 码 C++:
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: int m_isBalanced(TreeNode* root){ if(root==NULL) return 0; if(root->left==NULL && right==NULL) return 1; int left,right; left = m_isBalanced(root->left); right = m_isBalanced(root->right); if(left==INT_MAX || right==INT_MAX) return INT_MAX; if(left-right>1 || left-right<-1) return INT_MAX; return left>right? left+1:right+1; } bool isBalanced(TreeNode* root) { if(root==NULL) return true; if(root->left==NULL && right==NULL) return true; if(m_isBalanced(root)==INT_MAX) return false; return true; } };
运行结果:
执行用时 : 8 ms 内存消耗 : 16.7 MB