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?

这道题目最简单的方法是用recursive 做，其次是用 stack＋queue， 最难想清楚的是stack。当然写过了 也就不觉得难了。

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<Integer> postorderTraversal(TreeNode root) {        List<Integer> res = new ArrayList<Integer>();        if (root == null) {            return res;        }        Stack<TreeNode> st = new Stack<TreeNode>();        TreeNode pre = null;//用来判断 cur 与 pop 出的节点，或stack 中临近节点的关系        st.push(root);        while (!st.empty()) {            // cur always equals to st.peet();            TreeNode cur = st.peek();            // 说明只放了root， 或者pre 是cur 在stack 下面的元素，即说明还是在遍历左右子树的过程            if (pre == null || pre.left == cur || pre.right == cur) {                if (cur.left != null) {                    st.push(cur.left);                } else if (cur.right != null) {                    st.push(cur.right);                } else {                    res.add(cur.val);                    st.pop();                }            } else if (pre == cur.left) {               //则要判断cur 是否还有右节点，有的话，右节点入栈               if (cur.right != null) {                   st.push(cur.right);               } else {                   res.add(cur.val);                   st.pop();               }            } else {//即pre ==cur.right,此时说明 cur 的左右节点都已经遍历                res.add(cur.val);                st.pop();            }            pre = cur;        }        return res;    }}