【一天一道LeetCode】#94. Binary Tree Inorder Traversal
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(一)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].
(二)解题
题目大意:给定一个二叉树,输出中序遍历
讲二叉树的时候基本上都讲过递归求解中序遍历。即先访问左子树, 再根结点,再右子树,
/** * 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> ret; inorder(root,ret); return ret; } void inorder(TreeNode* p,vector<int>& ret) { if(p==NULL) return; inorder(p->left,ret);//访问左子树 ret.push_back(p->val);//将根节点保存 inorder(p->right,ret);//访问右子树 }};
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