Lintcode： 二叉树的后序遍历

问题：

给出一棵二叉树，返回其节点值的后序遍历。

样例：

样例 1:

输入：{1,2,3}
输出：[2,3,1]
解释： 
   1
  / \
 2   3
它将被序列化为{1,2,3}
后序遍历

样例 2:

输入：{1,#,2,3}
输出：[3,2,1]
解释： 
1
 \
  2
 /
3
它将被序列化为{1,#,2,3}
后序遍历

python：
 

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param root: A Tree
    @return: Postorder in ArrayList which contains node values.
    """
    def postorderTraversal(self, root):
        # write your code here
        endnode = False
        node = [root]
        result = []
        if root == None:
            return result
        while not endnode:
            while root.left != None or root.right != None:
                if root.left != None:
                    node.append(root)
                    root = root.left
                else:
                    node.append(root)
                    root = root.right
            supnode = node[-1]
            while root == supnode.right or supnode.right == None:
                result.append(root.val)
                del node[-1]
                if len(node) == 1:
                    result.append(supnode.val)
                    endnode = True
                    break
                root = supnode
                supnode = node[-1]
            if root == supnode.left and supnode.right != None:
                result.append(root.val)
                root = supnode.right
        return result

 

C++：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: A Tree
     * @return: Postorder in ArrayList which contains node values.
     */
    vector postorderTraversal(TreeNode * root) {
        // write your code here
        vector result;
        if(root == NULL)
        {
            return result;
        }
        
        vector node;
        TreeNode * temp = root;
        bool endnode = false;
        while(!endnode)
        {
            while(root->left != NULL || root->right != NULL)
            {
                if(root->left != NULL)
                {
                    node.push_back(root);
                    root = root->left;
                }else{
                    node.push_back(root);
                    root = root->right;
                }
            }
            TreeNode *supnode = node.back();
            
            while(root == supnode->right || supnode->right == NULL)
            {
                result.push_back(root->val);
                node.pop_back();
                if(node.size() == 0)
                {
                    result.push_back(supnode->val);
                    endnode = true;
                    break;
                }
                root = supnode;
                supnode = node.back();
            }
            
            if(root == supnode->left && supnode->right != NULL)
            {
                result.push_back(root->val);
                root = supnode->right;
            }

        }
        return result;
    }
};

 

PS：后序最难写，因为要判断节点在左支还是右支