二叉树的遍历,不管是recursive版的还是iterative版的,应该都是要求熟练掌握。
题目一:Binary Tree Postorder Traversal
Given a binary tree, return the postorder traversal of its nodes' value.
思路1:用递归的方法。
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> postorderTraversal(TreeNode *root) {
vector<int> res;
if(root==NULL) return res;
postorderTraversal(res, root); //如果不用这种引用的方式传参<span style="font-family:Comic Sans MS;">,那么设成全局遍历</span>
return res;
}
void postorderTraversal(vector<int> &res, TreeNode *root){
if(root==NULL) return;
postorderTraversal(res, root->left);
postorderTraversal(res, root->right);
res.push_back(root->val);
}
};
思路二:利用迭代的方式。迭代的方法有多种,其中使用two stack的效率较高点。利用栈后进先出的特点,将实现递归的效果,然后利用第二个栈,实现后续遍历的效果。
class Solution {
public:
vector<int> postorderTraversal(TreeNode *root) {
stack<TreeNode*> s;
stack<TreeNode*> output;
vector<int> v;
if(root==NULL) return v;
s.push(root);
while(!s.empty()){
TreeNode *node=s.top();
output.push(node);
s.pop();
if(node->left) s.push(node->left); //先左边,再右边
if(node->right) s.push(node->right);
}
while(!output.empty()){
v.push_back(output.top()->val);
output.pop();
}
return v;
}
};
题目二:Binary Tree Preorder Traversal
Given a binary tree, return the preorder traversal of its nodes' values
思路1:用递归的方法,使用全局量
class Solution {
private:
vector<int> res;
public:
vector<int> preorderTraversal(TreeNode *root) {
if(root==NULL) return res;
res.push_back(root->val);
preorderTraversal(root->left);
preorderTraversal(root->right);
return res;
}
};
思路二:使用iterative的方式,同样利用stack的特性实现递归。
class Solution {
public:
vector<int> preorderTraversal(TreeNode *root) {
stack<TreeNode*> s;
vector<int> v;
if(root==NULL) return v;
s.push(root);
while(!s.empty()){
TreeNode *node=s.top();
s.pop();
v.push_back(node->val);
if(node->right) s.push(node->right); //先右边,再左边
if(node->left) s.push(node->left);
}
return v;
}
};
题目三: Binary Tree Inorder Traversal**
Given a binary tree, return the inorder traversal of its nodes' values
思路:中序遍历的iterative方式和其他两种略有不同。先让所有左孩子都进栈,然后再一次让右孩子进栈。
class Solution {
public:
vector<int> inorderTraversal(TreeNode *root) {
stack<TreeNode*> nodes;
vector<int> res;
if(root==NULL) return res;
TreeNode *cur=root;
while(!nodes.empty() || cur){
while(cur){ //先一直走到最小面,并且进栈
nodes.push(cur);
cur=cur->left;
}
cur=nodes.top(); //走到头后开始退栈
nodes.pop();
res.push_back(cur->val);
cur=cur->right;
}
return res;
}
};