迭代解法在那篇面试总结二中写了
package Level3;
import java.util.ArrayList;
import Utility.TreeNode;
/**
* 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?
*
*/
public class S134 {
public static void main(String[] args) {
}
public ArrayList<Integer> postorderTraversal(TreeNode root) {
ArrayList<Integer> ret = new ArrayList<Integer>();
if(root == null){
return ret;
}
rec(root, ret);
return ret;
}
public void rec(TreeNode root, ArrayList<Integer> ret){
if(root == null){
return;
}
rec(root.left, ret);
rec(root.right, ret);
ret.add(root.val);
}
}
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public ArrayList<Integer> postorderTraversal(TreeNode root) {
ArrayList<Integer> ret = new ArrayList<Integer>();
if(root == null){
return ret;
}
Stack<TreeNode> s1 = new Stack<TreeNode>();
Stack<TreeNode> s2 = new Stack<TreeNode>();
s1.push(root);
while(!s1.isEmpty()){
TreeNode cur = s1.pop();
s2.push(cur);
if(cur.left != null){
s1.push(cur.left);
}
if(cur.right != null){
s1.push(cur.right);
}
}
while(!s2.isEmpty()){
ret.add(s2.pop().val);
}
return ret;
}
}