94. Binary Tree Inorder Traversal

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].

迭代版本

/**
 * 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) {
        stack<TreeNode *> st;
        vector<int> res;
        while(!st.empty() || root != NULL){
            if(root != NULL){
                st.push(root);
                root = root->left;
            }else{
                root = st.top();
                st.pop();
                res.push_back(root->val);
                root = root->right;
            }
        }
        return res;
    }
};

递归版本

/**
 * 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:
    //res为全局的
    vector<int> res;
    vector<int> inorderTraversal(TreeNode* root) {
        if(root == NULL)    return res;

        inorderTraversal(root->left);
        res.push_back(root->val);
        inorderTraversal(root->right);

        return res;
    }
};

另一种版本，res为函数局部变量

/**
 * 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> res;
        dfs(root, res);

        return res;
    }
private:
    void dfs(TreeNode *r, vector<int> &ret){
        if(r == NULL)   return;

        dfs(r->left, ret);
        ret.push_back(r->val);
        dfs(r->right, ret);
    }
};