[LeetCode] Binary Tree Inorder Traversal

[Problem]

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?

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

  1
 / \
2   3
   /
  4
   \
    5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

[Solution]

class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
       
        // result
        vector<int> res;
       
        // NULL node
        if(root == NULL)return res;
       
        // traversal the left brunch
        if(root->left != NULL){
            vector<int> left = inorderTraversal(root->left);
            vector<int>::iterator leftIt;
            for(leftIt = left.begin(); leftIt != left.end(); ++leftIt){
                res.push_back(*leftIt);
            }
        }
       
        // traversal the root
        res.push_back(root->val);
       
        // traversal the right brunch
        if(root->right != NULL){
            vector<int> right = inorderTraversal(root->right);
            vector<int>::iterator rightIt;
            for(rightIt = right.begin(); rightIt != right.end(); ++rightIt){
                res.push_back(*rightIt);
            }
        }
        return res;
    }
};

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