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68 · Binary Tree Postorder Traversal - LintCode

# Description

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

  • The first data is the root node, followed by the value of the left and right son nodes, and "#" indicates that there is no child node.
  • The number of nodes does not exceed 20.

Example 1:

Input:

binary tree = {1,2,3}

Output:

[2,3,1]

# Solution

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: A Tree
     * @return: Postorder in ArrayList which contains node values.
     */
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null){
            return result;
        }
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        TreeNode prev = null;
        
        while (!stack.isEmpty()){
            TreeNode curr = stack.peek();
            if (prev == null || prev.left == curr || prev.right == curr){
                if (curr.left != null){
                    stack.push(curr.left);
                } else if (curr.right != null){
                    stack.push(curr.right);
                }
            } else if (curr.left == prev){ // traverse up to root, see the right node
                if (curr.right != null){
                    stack.push(curr.right);
                }
            } else {
                result.add(curr.val);
                stack.pop();
            }
            prev = curr;
        }
        return result;
    }
}