Binary Tree Inorder Traversal——二叉樹的中序遍歷
阿新 • • 發佈:2019-01-04
原題:
Given a binary tree, return the inorder traversal of its nodes' values.
=>給定一個二叉樹,返回所有節點的中序遍歷
For example:
=>例如
Given binary tree {1,#,2,3}
,
=>給定二叉樹如下:
1 \ 2 / 3
return [1,3,2]
.
=>返回[1,3,2]
Note: Recursive solution is trivial, could you do it iteratively?
=>遞迴的演算法很簡單,能否不遞迴實現?
/**
* 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> inorderTraversal(TreeNode *root) {
}
};
曉東解析:
程式碼實現:
1)遞迴實現:
/** * 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> inorderTraversal(TreeNode *root) { vector<int> result; vector<int> left; vector<int> right; if(NULL == root) return result; left = inorderTraversal(root->left); if(left.size() != 0) result.insert(result.end(), left.begin(), left.end()); result.push_back(root->val); right = inorderTraversal(root->right); if(right.size() != 0) result.insert(result.end(), right.begin(), right.end()); return result; } };
執行結果:
67 / 67 test cases passed. | Status:
Accepted |
Runtime: 36 ms |
2)非遞迴實現
/**
* 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> inorderTraversal(TreeNode *root) {
vector<int> result;
stack<TreeNode*> TreeStack;
if(NULL == root) return result;
while(root || !TreeStack.empty()){
while(root){
TreeStack.push(root);
root = root->left;
}
root = TreeStack.top();
result.push_back(root->val);
TreeStack.pop();
root = root->right;
}
}
};
執行結果:
67 / 67 test cases passed. | Status:
Accepted |
Runtime: 36 ms |
若您有更好的演算法,歡迎提出指正。