Binary Tree Inorder Traversal

Description

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?

Solutions

Recursive

class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
        vector<int> res;
        inorder(root, res);
        return res;
    }
    
    void inorder(TreeNode *root, vector<int> &res) {
        if(root) {
            inorder(root->left, res);
            res.push_back(root->val);
            inorder(root->right, res);
        }
    }
};

Iterative

curr: the current node to be processed.

st: the stack of left sub-trees to be processed.

class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> res;
        stack<TreeNode *> st;
        TreeNode *curr = root, *node;
        while (curr || !st.empty()) {
            while (curr) {
                st.push(curr);
                curr = curr->left;
            }
            if (!st.empty()) {
                node = st.top();
                st.pop();
                res.push_back(node->val);
                curr = node->right;
            }
        }
        return res;
    }
};

One loop:

class Solution {
public:
    vector<int> inorderTraversal(TreeNode *root) {
        vector<int> res;
        stack<TreeNode *> st;
        TreeNode *curr = root, *node;
        while (true) {
            if (curr) {
                st.push(curr);
                curr = curr->left;
            } else if (!st.empty()) {
                node = st.top();
                st.pop();
                res.push_back(node->val);
                curr = node->right;
            } else {
                break;
            }
        }
        return res;
    }
};