Binary Tree Inorder Traversal

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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.

/**
 * 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) {
        stack<TreeNode *> nodestack;
        vector<int> arr(0);
        TreeNode *node;
        if(root == NULL) return arr;
        node = root;

        while(!nodestack.empty() || node!=NULL){
            if(node!=NULL) {
                nodestack.push(node);
                node=node->left;
            }
            else{
                node=nodestack.top();
                arr.push_back(node->val);
                nodestack.pop();
                node=node->right;
            }
        }
        return arr;
    }
};

Binary Tree Inorder Traversal

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原文地址：http://blog.csdn.net/uj_mosquito/article/details/42024429