[LintCode] 68 Binary Tree Postorder Traversal 解题报告

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



Example
Given binary tree {1,#,2,3},

   1
    \
     2
    /
   3


return [3,2,1].



Challenge
Can you do it without recursion?



思路
解法一：
Traverse


Code

/**
 * 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: The root of binary tree.
     * @return: Postorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        // write your code here
        
        ArrayList<Integer> result = new ArrayList<>();
        traverse(root, result);
        return result;
    }
    
    public void traverse(TreeNode root, ArrayList<Integer> result) {
        if (root == null) {
            return;
        }
        
        traverse(root.left, result);
        traverse(root.right, result);
        result.add(root.val);
    }
}

解法二：

Divide & Conquer

Code

/**
 * 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: The root of binary tree.
     * @return: Postorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        // write your code here
        
        ArrayList<Integer> result = new ArrayList<>();
        
        if (root == null) {
            return result;
        }
        
        ArrayList<Integer> left = postorderTraversal(root.left);
        ArrayList<Integer> right = postorderTraversal(root.right);
        
        result.addAll(left);
        result.addAll(right);
        result.add(root.val);
        
        return result;
    }
}

解法三：

Iterative

反方向pre-order。然后把答案reverse一下。


Code

/**
 * 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: The root of binary tree.
     * @return: Postorder in ArrayList which contains node values.
     */
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        // write your code here
        
        ArrayList<Integer> result = new ArrayList<>();
        
        if (root == null) {
            return result;
        }
        
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        
        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();
            result.add(node.val);
            
            if (node.left != null) {
                stack.push(node.left);
            }
            if (node.right != null) {
                stack.push(node.right);
            }
        }
        
        Collections.reverse(result);
        
        return result;
    }
}

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Stack<TreeNode> stack = new Stack<>();
        TreeNode prev = null;
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode node = stack.peek();
            if (prev == null || prev.left == node || prev.right == node) {
                if (node.left != null) {
                    stack.push(node.left);
                }
                else if (node.right != null) {
                    stack.push(node.right);
                }
            }
            else if (node.left == prev) {
                if (node.right != null) {
                    stack.push(node.right);
                }
            }
            else {
                result.add(node.val);
                stack.pop();
            }
            prev = node;
        }
        return result;
    }
}