Binary Tree Postorder Traversal

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

Example:

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

Output: [3,2,1]

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

Solution

1

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var postorderTraversal = function (root) {
    var res = [];
    helper(root, res);
    return res;
};

var helper = function (root, res) {
    if (!root) return;
    helper(root.left, res);
    helper(root.right, res);
    res.push(root.val);
};

Explain:

nope.

Complexity:

• Time complexity: O(n).
• Space complexity: O(n).

Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var postorderTraversal = function (root) {
    if (!root) return [];
    var res = [];
    var stack = [];
    var node = root;
    while (node || stack.length) {
        if (node) {
            stack.push(node);
            res.unshift(node.val);
            node = node.right;
        } else {
            node = stack.pop();
            node = node.left;
        }
    }
    return res;
};

Complexity:

• Time complexity: O(n).
• Space complexity: O(n).