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.
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isBalanced(TreeNode *root) {
int depth;
return checkBalanced(root, depth);
}
bool checkBalanced(TreeNode *root, int &depth) // 给checkBalanced传入depth是为了求depth的值,由于checkBalanced只能返回一个值,所以传入一个引用,求depth的值。
{
if(root==NULL)
{
depth=0;
return true;
}
int depthLeft,depthRight;
bool leftBalanced=checkBalanced(root->left, depthLeft);
bool rightBalanced=checkBalanced(root->right, depthRight);
depth=max(depthLeft+1,depthRight+1);
return leftBalanced&&rightBalanced&&(abs(depthLeft-depthRight)<=1);
}
};
本文介绍了一种用于判断二叉树是否为高度平衡的方法。高度平衡二叉树定义为对于树中每一个节点,其两个子树的深度相差不超过1。通过递归计算每个节点的左右子树深度来实现。
367

被折叠的 条评论
为什么被折叠?



