lintcode,二叉树的后序遍历

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给出一棵二叉树，返回其节点值的后序遍历。
样例
给出一棵二叉树 {1,#,2,3},
1
\
2
/
3
返回 [3,2,1]
挑战
你能使用非递归实现么？

解题思路：递归方法容易实现。非递归考虑用两个栈，类似栈实现队列的思路，将二叉树后序入栈。
一刷ac

/** * 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) {        ArrayList<Integer> rst = new ArrayList<Integer>();        if(root == null) return rst;        ArrayList<Integer> left = postorderTraversal(root.left);        ArrayList<Integer> right = postorderTraversal(root.right);        rst.addAll(left);        rst.addAll(right);        rst.add(root.val);        return rst;    }}

非递归两个栈。

/** * 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) {        ArrayList<Integer> res = new ArrayList<Integer>();        if(root == null) return res;        Stack<TreeNode> stack1 = new Stack<TreeNode>();        Stack<TreeNode> stack2 = new Stack<TreeNode>();        stack1.push(root);        while(!stack1.isEmpty()){            TreeNode node = stack1.pop();            stack2.push(node);            if(node.left != null)stack1.push(node.left);            if(node.right != null)stack1.push(node.right);        }        while(!stack2.isEmpty()){            res.add(stack2.pop().val);        }        return res;    }}

非递归一个栈

/** * 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) {        ArrayList<Integer> res = new ArrayList<Integer>();        if(root == null) return res;        Stack<TreeNode> stack = new Stack<TreeNode>();        stack.push(root);        while(!stack.empty()){            TreeNode node = stack.pop();            res.add(0,node.val);            if(node.left != null) stack.push(node.left);            if(node.right != null) stack.push(node.right);        }        return res;    }}

非递归一个栈

/** * 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) {        ArrayList<Integer> res = new ArrayList<Integer>();        if(root == null) return res;        Stack<TreeNode> stack = new Stack<TreeNode>();        while(root != null){            res.add(0,root.val);            if(root.left != null) stack.push(root.left);            root = root.right;            if(root == null && !stack.empty()){                root = stack.pop();            }        }        return res;    }}

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