94. Binary Tree Inorder Traversal

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

Example:

Input: [1,null,2,3]
   1
    
     2
    /
   3

Output: [1,3,2]

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

非递归二叉树中序遍历

java：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>() ;
        if (root == null)
            return res ;
        Stack<TreeNode> stack = new Stack<>() ;
        TreeNode cur = root ;
        while(cur != null || !stack.isEmpty()){
            while(cur != null){
                stack.push(cur) ;
                cur = cur.left ;
            }
            TreeNode node = stack.pop() ;
            res.add(node.val) ;
            cur  = node.right ;
        }
        return res ;
    }
}

C++:

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