110. Balanced Binary Tree
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题目:平衡二叉树
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.
题意:
给定一个二叉树,检测该树是否是高度平衡二叉树。
对于这个问题,一个高度平衡二叉树的定义为:每个节点的两个子树的深度不会相差超过1.
思路一:
自顶向下,递归实现,二叉树是否平衡,根据题目中的定义,高度平衡二叉树是每一个节点的两个字数的深度差不能超过1,那么我们肯定需要一个求各个点深度的函数,然后对每个节点的两个子树来比较深度差,时间复杂度为O(NlgN)。
代码:C++版:16ms
/** * 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: bool isBalanced(TreeNode* root) { if (!root) return true; if (abs(getDepth(root->left) - getDepth(root->right)) > 1) return false; //左右子树深相差大于1,则返回false return isBalanced(root->left) && isBalanced(root->right); //左右子树均要高度平衡 } int getDepth(TreeNode *root) { //计算树深 if (!root) return 0; return 1 + max(getDepth(root->left), getDepth(root->right)); }};
思路二:
思路一思路比较直观,但是不是太高效,每次每个节点均需要访问计算深度一次,优化如下,自底向上,通过递归实现,得到每一次树深的值,判断两个子树深度相差是否大于1,大于1说明该二叉树不是高度平衡的二叉树,则返回-1代表不是高度平衡二叉树。如果返回的树深大于等于0,则是高度平衡二叉树。
代码:C++版:16ms
/** * 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: bool isBalanced(TreeNode* root) { return balancedHeight(root) >= 0; } int balancedHeight(TreeNode *root) { if (root == nullptr) return 0; //终止条件 int left = balancedHeight(root->left); int right = balancedHeight(root->right); if (left < 0 || right < 0 || abs(left - right) > 1) return -1; //减枝 return max(left, right) + 1; //三方合并 }};
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