binary-tree-postorder-traversal[后序遍历]

题目描述

Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree{1,#,2,3},

return[3,2,1].
Note: Recursive solution is trivial, could you do it iteratively?

• 思路
  求后序遍历。
  法一：递归（参考中序遍历）
  法二：非递归。判断栈顶节点，若无孩子节点或其孩子节点都被访问（将右左孩子依次入栈，保证了根节点始终在子节点之后被访问），则访问该节点，否则把该节点的左右孩子加入栈中（注意顺序）。

//非递归
/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.*;
public class Solution {
    public ArrayList postorderTraversal(TreeNode root) {
        ArrayList list = new ArrayList();
        if(root == null)
            return list;
        
        Stack stack = new Stack();
        stack.add(root);
        TreeNode pre = null; //前一个访问的节点
        
        while( !stack.isEmpty()){
            TreeNode node = stack.peek();
            //当是根节点或其子节点都被访问时
            if((node.left == null && node.right == null) ||
              (pre != null && (pre == node.left || pre == node.right))){
                list.add(node.val);
                stack.pop();
                pre = node;
            }else{
                if(node.right != null)
                    stack.add(node.right);
                if(node.left != null)
                    stack.add(node.left);
            }
        }
        
        return list;
    }
}