Lintcode 二叉樹的後序遍歷

二叉樹的後序遍歷

給出一棵二叉樹，返回其節點值的後序遍歷。

樣例

給出一棵二叉樹 {1,#,2,3},

   1
    \
     2
    /
   3

返回 [3,2,1]

二叉樹後序遍歷：左子樹->右子樹->根節點。下圖二叉樹的後序遍歷結果爲：edgfbca。




通過遞歸方式實現二叉樹後序遍歷的代碼如下：

/**
 * 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.
     */  
    ArrayList<Integer> list = new ArrayList<Integer>();
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        // write your code here
        postorder(root);
        return list;
    }
    private void postorder(TreeNode root){
        if(root == null){
            return;
        }
        postorder(root.left);
        postorder(root.right);
        list.add(root.val);
    }
}

通過非遞歸方式實現二叉樹的後序遍歷的代碼如下：

/**
 * 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.
     */  
     ArrayList<Integer> list = new ArrayList<Integer>();
     Stack<TreeNode> stack = new Stack<TreeNode>();
    public ArrayList<Integer> postorderTraversal(TreeNode root) {
        // write your code here
        postorder(root);
        return list;
    }
    private void postorder(TreeNode root){
        TreeNode p = root;
        while(root!=null){
            while(root.left!=null){
                stack.push(root);
                root = root.left;
            }
            while(root!=null&&(root.right==null||root.right == p)){
                list.add(root.val);
                p = root;
                if(stack.isEmpty()) return;
                root = stack.pop();
            }
            stack.push(root);
            root = root.right;
        }
    }
}