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Programmingoneonone
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Leetcode Binary Tree Inorder Traversal problem solution

YASH PAL, 31 July 202419 January 2026

In this Leetcode Binary Tree Inorder Traversal problem solution, we have given the root of a binary tree, return the inorder traversal of its nodes’ values.

Leetcode Binary Tree Inorder Traversal problem solution

Leetcode Binary Tree Inorder Traversal problem solution in Python.

from collections import deque

class Solution(object):
    def _in_order_iter(self, root):
        stack = deque()
        while stack or root:
            if root:
                stack.append(root)
                root = root.left
            else:
                root = stack.pop()
                yield root.val
                root = root.right

    def inorderTraversal(self, root):
        return list(self._in_order_iter(root))

Binary Tree Inorder Traversal problem solution in Java.

public List<Integer> inorderTraversal(TreeNode root) {
    List<Integer> resultList = new ArrayList<Integer>();
    Stack<TreeNode> stack = new Stack<TreeNode>();
	TreeNode current = root; 
    while (current != null || !stack.isEmpty()) {
        while (current != null) { 
            stack.push(current);
            current = current.left;
        }
        current = stack.pop();
        resultList.add(current.val);
        current = current.right;
    }
    return resultList;
}

Problem solution in C++.

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

Problem solution in C.

int *ans;
int cnt;
int base;
void tra(struct TreeNode *node){
    if(node==NULL){
        return;
    }
    tra(node->left);
    if(cnt==base){
        //not enough memory
        base*=2;
        ans=realloc(ans,base*sizeof(int));
    }
    ans[cnt++]=node->val;
    tra(node->right);
}
int* inorderTraversal(struct TreeNode* root, int* rs){
    base=8;
    ans=malloc(sizeof(int)*base);
    cnt=0;
    tra(root);
    *rs=cnt;
    return ans;
}

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