平衡二叉树判断 完全二叉树判断
平衡二叉树判断 有一棵二叉树，请设计一个算法判断这棵二叉树是否为平衡二叉树。 给定二叉树的根结点root，请返回一个bool值，代表这棵树是否为平衡二叉树。 我的提交
-- coding:utf-8 --
class TreeNode:
def init(self, x):
self.val = x
self.left = None
self.right = None
class CheckBalance: def getDepth(self, root): if not root: return 0 ldepth = self.getDepth(root.left) rdepth = self.getDepth(root.right) return max(ldepth, rdepth) + 1
def check(self, root):
# write code here
if not root:
return True
ldepth = self.getDepth(root.left)
rdepth = self.getDepth(root.right)
diff = abs(ldepth - rdepth)
if diff > 1:
return False
return self.check(root.left) and self.check(root.right)
参考答案 /* struct TreeNode { int val; struct TreeNode *left; struct TreeNode right; TreeNode(int x) : val(x), left(NULL), right(NULL) { } };/
class CheckBalance { public: int chk(TreeNode* rt){ if(rt == NULL) return 0; int l = chk(rt->left),r = chk(rt->right); if(l < 0 || r < 0) return -1; if(abs(r - l) > 1) return -1; return r > l ? r + 1 : l + 1; } bool check(TreeNode* root) { return chk(root) >= 0; } }; 完全二叉树判断 有一棵二叉树,请设计一个算法判断它是否是完全二叉树。 给定二叉树的根结点root，请返回一个bool值代表它是否为完全二叉树。树的结点个数小于等于500 我的提交
-- coding:utf-8 --
class TreeNode:
def init(self, x):
self.val = x
self.left = None
self.right = None
class CheckCompletion: def chk(self, root): # write code here if not root: return True queue = [] stop = False queue.append(root) while queue: p = queue.pop(0) if p.left: if stop: return False queue.append(p.left) else: stop = True if p.right: if stop: return False queue.append(p.right) else: stop = True return True 参考答案 /* struct TreeNode { int val; struct TreeNode *left; struct TreeNode right; TreeNode(int x) : val(x), left(NULL), right(NULL) { } };/
class CheckCompletion {
public:
bool chk(TreeNode* root) {
TreeNode *q[800],*now;
int head = 0,tail = 0;
q[tail ++] = root;
while(head != tail){
now = q[head ++];
if(now->left) q[tail ++] = now->left;
if(now->right) q[tail ++] = now->right;
}
bool flag = false;
for(int i = 0;i < tail;++ i)
if(q[i]->left == NULL || q[i]->right == NULL){
if(q[i]->right && q[i]->left == NULL) return false;
if(!flag) flag = true;
else if(q[i]->left || q[i]->right) return false;
}
return true;
}
};
