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?

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);
	}
}

LeetCode | Binary Tree Postorder Traversal

原文：http://blog.csdn.net/perfect8886/article/details/19014737