对称二叉树
递归写法:
class Solution {
public:
bool isSymmetrical(TreeNode* pRoot) {
if (pRoot == NULL) {
return true;
}
return IsSymmetrical(pRoot->left, pRoot->right);
}
bool IsSymmetrical(TreeNode* lt, TreeNode* rt) {
if (lt == NULL && rt == NULL) {
return true;
}
if (lt == NULL || rt == NULL) {
return false;
}
return ( lt->val == rt->val && IsSymmetrical(lt->left, rt->right) && IsSymmetrical(lt->right, rt->left) );
}
};
非递归写法:
class Solution {
public:
bool isSymmetrical(TreeNode* pRoot) {
if (pRoot == NULL) {
return true;
}
if (pRoot->left == NULL && pRoot->right == NULL) {
return true;
}
if (pRoot->left == NULL || pRoot->right == NULL) {
return false;
}
stack<TreeNode*> ltree;
stack<TreeNode*> rtree;
// 非递归遍历左、右子树
// 左子树按根左右的顺序遍历,右子树按根右左的顺序遍历
// 注意NULL节点也要存入栈中(特例:一棵树所有节点值都相等)
ltree.push(pRoot->left);
rtree.push(pRoot->right);
while (!ltree.empty() && !rtree.empty()) {
// 左子树
TreeNode* cur1 = ltree.top(); ltree.pop();
if (cur1 != NULL) {
ltree.push(cur1->right);
ltree.push(cur1->left);
}
// 右子树
TreeNode* cur2 = rtree.top(); rtree.pop();
if (cur2 != NULL) {
rtree.push(cur2->left);
rtree.push(cur2->right);
}
// 对称的话,应该取出的节点的值相等
if (cur1 == NULL && cur2 == NULL) continue;
if (cur1 == NULL || cur2 == NULL) return false;
if (cur1->val != cur2->val) return false;
}
return ( (ltree.empty() && rtree.empty()) ? true : false);
}
};