leetcode: Binary Tree Postorder Traversal

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Given a binary tree, return the inorder traversal of its nodes‘ values.

For example:
Given binary tree {1,#,2,3},

   1
         2
    /
   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

题目描述：

二叉树的中序遍历。先遍历左子树，再遍历根节点，再再遍历右子树。

代码实现：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> inorderTraversal(TreeNode* root) {
    vector<int> re;
    const TreeNode *p;
    p=root;
    stack<const TreeNode *>s;
    while(!s.empty() || p!=NULL)
    {
        if(p!=NULL)
        {
            s.push(p);
            p=p->left;
        }
        else
        {
            p=s.top();
            s.pop();
            re.push_back(p->val);
            p=p->right;
        }
    }
    return re;
    }
};

leetcode: Binary Tree Postorder Traversal

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原文地址：http://blog.csdn.net/flyljg/article/details/48006745