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# 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);
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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