94. Binary Tree Inorder Traversal
Given a binary tree, return the inorder traversal of its nodes' values.
Example:
Input: [1,null,2,3]
1
\
2
/
3
Output: [1,3,2]
Follow up: Recursive solution is trivial, could you do it iteratively?
题意:二叉树的中序遍历,要求用迭代
思路:使用到栈,将整棵树的最左边的一条链压入栈中,每次取出栈顶元素,如果它有右子树,则将右子树压入栈中。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int>res;
stack<TreeNode*> stk;
auto p=root;
while(p||stk.size())
{
while(p)//整颗子树的最左侧进栈
{
stk.push(p);
p=p->left;
}
p=stk.top();
stk.pop();
res.push_back(p->val);
p=p->right;
}
return res;
}
};
给出递归写法
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<int> inorderTraversal(TreeNode* root) {
vector<int> result;
inorder(root,result);
return result;
}
void inorder(TreeNode* root,vector<int>& v){
if(root!=nullptr){
inorder(root->left,v);
v.push_back(root->val);
inorder(root->right,v);
}
}
};
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