LeetCode: Binary Tree Inorder Traversal

一次过

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void dfs(TreeNode *root, vector<int> &ret) {
        if (!root) return;
        dfs(root->left, ret);
        ret.push_back(root->val);
        dfs(root->right, ret);
    }
    vector<int> inorderTraversal(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<int> ret;
        dfs(root, ret);
        return ret;
    }
};

 写了一个不需要递归的代码，这段代码更好

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        stack<TreeNode *> S;
        vector<int> ret;
        if (!root) return ret;
        S.push(root);
        while (!S.empty()) {
            TreeNode *tmp = S.top();
            if (tmp->left) {
                S.push(tmp->left);
                tmp->left = NULL;
            }
            else {
                ret.push_back(tmp->val);
                S.pop();
                if (tmp->right) S.push(tmp->right);
            }
        }
        return ret;
    }
};

 C#

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<int> InorderTraversal(TreeNode root) {
        Stack<TreeNode> S = new Stack<TreeNode>();
        List<int> ans = new List<int>();
        if (root == null) return ans;
        S.Push(root);
        while(S.Count != 0)
        {
            TreeNode top = S.Peek();
            if (top.left != null) {
                S.Push(top.left);
                top.left = null;
            }
            else {
                ans.Add(top.val);
                S.Pop();
                if (top.right != null) S.Push(top.right);
            }
        }
        return ans;
    }
}