树其实是递归定义的,所以递归遍历树的方式也是比较自然的。但是因为种种原因---比如面试官想测你智商---你必须采用非递归的方法遍历树,那么请记住以下两个诀窍:
1. 树的非递归遍历是对递归遍历的模仿,而怎么模仿递归呢?使用栈和循环
2. 都只需要一个循环
下面是code,都通过了leetcode的测试:
-
/*前序遍历*/
-
vector<int> preorderTraversal(TreeNode *root){
-
vector<int> result;
-
if(root!=NULL){
-
stack<TreeNode *> s;
-
bool done=false;
-
TreeNode * current = root;
-
while(!done){
-
if(current!=NULL){
-
s.push(current);
-
result.push_back(current->val);
-
current=current->left;
-
}
-
else{
-
if(s.empty()){
-
done=true;
-
}
-
else{
-
current=s.top();
-
s.pop();
-
current=current->right;
-
}
-
}
-
}
-
}
-
return result;
-
}
-
/*中序遍历*/
-
vector<int> inorderTraversal(TreeNode *root){
-
vector<int> result;
-
if(root!=NULL){
-
stack<TreeNode *> s;
-
bool done=false;
-
TreeNode * current = root;
-
while(!done){
-
if(current!=NULL){
-
s.push(current);
-
-
current=current->left;
-
}
-
else{
-
if(s.empty()){
-
done=true;
-
}
-
else{
-
current=s.top();
-
result.push_back(current->val);
-
s.pop();
-
current=current->right;
-
}
-
}
-
}
-
}
-
return result;
-
}
-
/*后序遍历*/
-
vector<int> postorderTraversal(TreeNode *root) {
-
stack<TreeNode *> s;
-
vector<int> result;
-
if(root!=NULL){
-
s.push(root);
-
while(!s.empty()){
-
TreeNode * current=s.top();
-
s.pop();
-
result.push_back(current->val);
-
if(current->left!=NULL)
-
s.push(current->left);
-
if(current->right!=NULL)
-
s.push(current->right);
-
}
-
}
-
reverse(result.begin(), result.end());
-
return result;
-
}
-
阅读(725) | 评论(0) | 转发(0) |