145.Binary Tree Postorder Traversal

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Question

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

Example:

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

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

Left, Right, root

Answer

Recursive
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/**
 * 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<>();
        return recursivePostTraversal(root, res);
    }
    public List<Integer> recursivePostTraversal(TreeNode root, List<Integer> res){
        if(root!=null){
            recursivePostTraversal(root.left, res);
            recursivePostTraversal(root.right, res);
            res.add(root.val);
        }
        return res;
    }
}

Time complexity: O(n), 1ms

Iterative, Using addFirst()
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/**
 * 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) {
        LinkedList<Integer> res = new LinkedList<>();
        if(root == null)
            return res;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while(!stack.isEmpty()){
            TreeNode node = stack.pop();
            res.addFirst(node.val);
            if(node.left!=null)
                stack.push(node.left);
            if(node.right!=null)
                stack.push(node.right);
        }
        return res;
    }
}

Time complexity: O(n), 2ms

push node.left first, then node.right, in this way, will addFirst(node.right), then addFirst(add.left) -> left, right, root

Only LinkedList has method addFirst(Element e)

Iterative, Using LastVisit node
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/**
 * 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) {
        LinkedList<Integer> res = new LinkedList<>();
        if(root == null)
            return res;
        Stack<TreeNode> stack = new Stack<>();
        TreeNode node = root;
        TreeNode lastVisit = root;
        while(node != null || !stack.isEmpty()){
            while(node!=null){
                stack.push(node);
                node = node.left;
            }
            node = stack.peek();
            if(node.right==null || node.right==lastVisit){
                res.add(node.val);
                stack.pop();
                lastVisit = node;
                node = null;
            }
            else
                node = node.right;
        }
        return res;
    }
}

后序遍历在决定是否可以输出当前节点的值的时候，需要考虑其左右子树是否都已经遍历完成。

所以需要设置一个lastVisit游标。

若lastVisit等于当前考查节点的右子树，表示该节点的左右子树都已经遍历完成，则可以输出当前节点。

并把lastVisit节点设置成当前节点，将当前游标节点node设置为空，下一轮就可以访问栈顶元素。