LeetCode | Binary Tree Postorder Traversal

题目

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?
分析
题目说了，递归太简单，有本事整个非递归的。

二叉树的先序、中序、后序三种遍历方式中，后序遍历的非递归实现稍微复杂些，需要额外开辟空间记录每个节点的右孩子是否已经访问。

代码

import java.util.ArrayList;
import java.util.Stack;

public class BinaryTreePostorderTraversal {
public class TreeNode {
int val;
TreeNode left;
TreeNode right;

TreeNode(int x) {
val = x;
}
}

public ArrayList<Integer> postorderTraversal(TreeNode root) {
ArrayList<Integer> list = new ArrayList<Integer>();
// recursivePostorderTraversal(root, list);
iterativePostorderTraversal(root, list);
return list;

}

class TreeNodeWithFlag {
TreeNode node;
boolean isVisited;

TreeNodeWithFlag(TreeNode node, boolean isVisited) {
this.node = node;
this.isVisited = isVisited;
}
}

private void iterativePostorderTraversal(TreeNode root,
ArrayList<Integer> list) {
if (root == null) {
return;
}

Stack<TreeNodeWithFlag> stack = new Stack<TreeNodeWithFlag>();
TreeNode node = root;
while (node != null) {
stack.push(new TreeNodeWithFlag(node, false));
node = node.left;
}
while (!stack.isEmpty()) {
TreeNodeWithFlag nodeWithFlag = stack.peek();
node = nodeWithFlag.node;
if (node.right == null || nodeWithFlag.isVisited) {
nodeWithFlag = stack.pop();
list.add(nodeWithFlag.node.val);
} else {
nodeWithFlag.isVisited = true;
node = node.right;
while (node != null) {
stack.push(new TreeNodeWithFlag(node, false));
node = node.left;
}
}
}
}

private void recursivePostorderTraversal(TreeNode root,
ArrayList<Integer> list) {
if (root == null) {
return;
}
recursivePostorderTraversal(root.left, list);
recursivePostorderTraversal(root.right, list);
list.add(root.val);
}
}