Leetcode Binary Tree Postorder Traversal(面试题推荐)

此题来自leetcode   https://oj.leetcode.com/problems/binary-tree-postorder-traversal/

Given a binary tree, return the postorder traversal of its nodes' values.

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

   1    \     2    /   3

return [3,2,1].

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

   非递后续归遍历二叉树，肯定得用到栈。先序遍历很好写，但后续遍历就不是那么容易了。

   只需要设置个指针pre，指向最后输出的那个节点就行了，只要判断cur指针指向的是上次输出节点的父节点，且cur无其他未遍历的节点，这个时候就把cur节点输出即可，然后更改pre。原理是要遍历当前节点，其所有子节点都必须遍历完，因为肯定是先左后右，所以只需一个指针保持前一次输出的结果即可。

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