Binary Tree Postorder Traversal LeetCode Solution

Problem – Binary Tree Postorder Traversal LeetCode Solution

Given the root of a binary tree, return the postorder traversal of its nodes’ values.

Example 1:

Input: root = [1,null,2,3]
Output: [3,2,1]

Example 2:

Input: root = []
Output: []

Example 3:

Input: root = [1]
Output: [1]

Constraints:

• The number of the nodes in the tree is in the range [0, 100].
• -100 <= Node.val <= 100

Binary Tree Postorder Traversal LeetCode Solution in Java

public List<Integer> preorderTraversal(TreeNode root) {
    List<Integer> result = new ArrayList<>();
    Deque<TreeNode> stack = new ArrayDeque<>();
    TreeNode p = root;
    while(!stack.isEmpty() || p != null) {
        if(p != null) {
            stack.push(p);
            result.add(p.val);  // Add before going to children
            p = p.left;
        } else {
            TreeNode node = stack.pop();
            p = node.right;   
        }
    }
    return result;
}

Binary Tree Postorder Traversal LeetCode Solution in Python

class Solution:
    # @param {TreeNode} root
    # @return {integer[]}
    def postorderTraversal(self, root):
        traversal, stack = [], [(root, False)]
        while stack:
            node, visited = stack.pop()
            if node:
                if visited:
                    # add to result if visited
                    traversal.append(node.val)
                else:
                    # post-order
                    stack.append((node, True))
                    stack.append((node.right, False))
                    stack.append((node.left, False))

        return traversal

Binary Tree Postorder Traversal LeetCode Solution in C++

class Solution {
public:
    vector<int> postorderTraversal(TreeNode* root) {
        vector<int> nodes;
        stack<TreeNode*> todo;
        TreeNode* last = NULL;
        while (root || !todo.empty()) {
            if (root) {
                todo.push(root);
                root = root -> left;
            } else {
                TreeNode* node = todo.top();
                if (node -> right && last != node -> right) {
                    root = node -> right;
                } else {
                    nodes.push_back(node -> val);
                    last = node;
                    todo.pop();
                }
            }
        }
        return nodes;
    }
};

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