Leetcode 树 Binary Tree Inorder Traversal

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Binary Tree Inorder Traversal

Total Accepted: 16984 Total
Submissions: 48800

Given a binary tree, return the inorder traversal of its nodes‘ values.

For example:

Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

confused what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.

题意：中序遍历

思路：采用递归实现。因为函数声明是返回一个vector<int>,所以每个子树返回的是该子树的中序遍历的结果

按照 左、根、右的次序把根和左右子树的vector合并起来就可以了

复杂度：时间O(n)，空间O(n)

相关题目：

Binary Tree Preorder Traversal

Binary Tree Inorder Traversal

/**
 * 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> in;
    	if(root == NULL) return in;
    	TreeNode *left = root->left;
    	TreeNode *right = root->right;

    	if(left) {
    		vector<int> left_vector = inorderTraversal(left);
    		in.insert(in.end(), left_vector.begin(), left_vector.end());
    	}

    	in.push_back(root->val);

    	if(right){
    		vector<int> right_vector = inorderTraversal(right);
    		in.insert(in.end(), right_vector.begin(), right_vector.end());
    	}

    	return in;
    }
};

Leetcode 树 Binary Tree Inorder Traversal