题目:
Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree [1,null,2,3],
1
\
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
递归采取中序遍历即可:
/**
* 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:
void inorder(TreeNode* root, vector<int>& result) {
if (!root) return;
inorder(root -> left, result);
result.push_back(root -> val);
inorder(root -> right, result);
}
vector<int> inorderTraversal(TreeNode* root) {
vector<int> result;
inorder(root, result);
return result;
}
};
本文介绍了一种实现二叉树中序遍历的方法,通过递归方式获取节点值并返回结果。示例中使用了一个包含三个节点的二叉树,并给出了相应的中序遍历结果。
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