Morris Traversal 二叉树遍历。

那天做了个SWAP NODE的题，要求constant space，不得不Morris Traversal。

稍微研究了一下，真正意义上的O(1)space对二叉树进行遍历。好像是1979年的算法。

第一次看着挺乱的，智商严重不足，不得不在纸上画出来，一目了然。。。建议大家自己动手画一下。

每个没有右节点的Node要建1个辅助path，回到他来自的那个左节点的parent。很绕口= =然后第二次来到这个点的时候，要去掉这个Path。所以每个点都遍历了2次，时间上还是O(n)..

和那个树状数组有点像，以每个左节点为中心。

import java.util.*;

public class Morris {
	public static class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;

		public TreeNode (int v) {
			this.val = v;
			left = null;
			right = null;
		}
	}


		public static List<Integer> preOrder(TreeNode root) {
			List<Integer> res = new ArrayList<>();
			TreeNode temp = root;
			while (temp != null) {
				 TreeNode morisTemp = temp;
				 if (temp.left != null) {
				 	morisTemp = temp.left;

				 	while (morisTemp.right != null && morisTemp.right != temp) {
				 		morisTemp = morisTemp.right;
				 	}

				 	// first time, need to add a path
				 	if (morisTemp.right == null) {
				 		morisTemp.right = temp;
				 		//res.add(temp.val);	//pre-Order
				 		temp = temp.left;
				 	} else {	// second time, remove the path we added,
				 		morisTemp.right = null;
				 		//res.add(temp.val);	//in-Order
				 		temp = temp.right;
				 	}

				 } else {
				 	res.add(temp.val);
				 	temp = temp.right;
				 }
			}

			return res;
		}
	

	public static void main(String[] args) {
		TreeNode n10 = new TreeNode(10);
		TreeNode n4 = new TreeNode(4);
		TreeNode n1 = new TreeNode(1);
		TreeNode n5 = new TreeNode(5);
		TreeNode n3 = new TreeNode(3);
		TreeNode n6 = new TreeNode(6);
		TreeNode n12 = new TreeNode(12);
		TreeNode n11 = new TreeNode(11);
		TreeNode n14 = new TreeNode(14);
		TreeNode n15 = new TreeNode(15);

		n10.left = n4;
		n10.right = n12;
		n4.left = n1;
		n4.right = n5;
		n5.left = n3;
		n5.right = n6;
		n12.left = n11;
		n12.right = n14;
		n14.left = n15;

		for (int i : preOrder(n10)) {
			System.out.print(i + " ");
		}
	}

}

preOrder和inOrder做出来了，postOrder有点麻烦，不会做= =再研究吧。。