145. 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?

 

M1: recursive

time: O(n), space: O(h)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        postorder(root, res);
        return res;
    }
    
    private void postorder(TreeNode root, List<Integer> res) {
        if(root == null) {
            return;
        }
        postorder(root.left, res);
        postorder(root.right, res);
        res.add(root.val);
    }
}

M2: iterative

Let's start from the root and then at each iteration pop the current node out of the stack and push its child nodes. In the implemented strategy we push nodes into stack following the order Top->Bottom and Left->Right. Since DFS postorder transversal is Bottom->Top and Left->Right the output list should be reverted after the end of loop.

和preorder traversal的iterative解法相似。preorder traversal的iterative解法先从stack中弹出当前node并加入res中，如果当前node有右子节点，右子节点入栈，如果有左子节点，左子节点入栈（stack后进先出）

但由于preorder traversal初始时也是把root入栈，先弹出就必须加在res的末尾，所以要用res.add(0, node.val)，所以添加node子节点的顺序有调整：如果当前node有左子节点，左子节点入栈；如果当前node有右子节点，右子节点入栈。弹出的时候先弹出右子节点，加在root的左边，再弹出左子节点，加在right的左边

time: O(n), space: O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        LinkedList<TreeNode> s = new LinkedList<>();
        if(root == null) {
            return res;
        }
        
        s.offerFirst(root);
        while(!s.isEmpty()) {
            TreeNode cur = s.pollFirst();
            res.add(0, cur.val);
            if(cur.left != null) {
                s.offerFirst(cur.left);
            }
            if(cur.right != null) {
                s.offerFirst(cur.right);
            }
        }
        return res;
    }
}