一种遍历二叉树的方式，并且时间复杂度O(N)，额外空间复杂度O(1)
通过利用原树中大量空闲指针的方式，达到节省空间的目的
Morris遍历细节
假设来到当前节点cur，开始时cur来到头节点位置
1）如果cur没有左孩子，cur向右移动(cur = cur.right)
2）如果cur有左孩子，找到左子树上最右的节点mostRight：
a.如果mostRight的右指针指向空，让其指向cur，然后cur向左移动(cur = cur.left)
b.如果mostRight的右指针指向cur，让其指向null，然后cur向右移动(cur = cur.right)
3）cur为空时遍历停止
Morris遍历实质
建立一种机制：
对于没有左子树的节点只到达一次，
对于有左子树的节点会到达两次
morris遍历时间复杂度依然是O(N)
习题1 Morris遍历
public static class Node {
public int value;
Node left;
Node right;
public Node(int data) {
this.value = data;
}
}
public static void process(Node root) {
if (root == null) {
return;
}
// 1
process(root.left);
// 2
process(root.right);
// 3
}
public static void morris(Node head) {
if (head == null) {
return;
}
Node cur = head;
Node mostRight = null;
while (cur != null) {
mostRight = cur.left;
if (mostRight != null) {
while (mostRight.right != null && mostRight.right != cur) {
mostRight = mostRight.right;
}
if (mostRight.right == null) {
mostRight.right = cur;
cur = cur.left;
continue;
} else {
mostRight.right = null;
}
}
cur = cur.right;
}
}
public static void morrisPre(Node head) {
if (head == null) {
return;
}
Node cur = head;
Node mostRight = null;
while (cur != null) {
mostRight = cur.left;
if (mostRight != null) {
while (mostRight.right != null && mostRight.right != cur) {
mostRight = mostRight.right;
}
if (mostRight.right == null) {
System.out.print(cur.value + " ");
mostRight.right = cur;
cur = cur.left;
continue;
} else {
mostRight.right = null;
}
} else {
System.out.print(cur.value + " ");
}
cur = cur.right;
}
System.out.println();
}
public static void morrisIn(Node head) {
if (head == null) {
return;
}
Node cur = head;
Node mostRight = null;
while (cur != null) {
mostRight = cur.left;
if (mostRight != null) {
while (mostRight.right != null && mostRight.right != cur) {
mostRight = mostRight.right;
}
if (mostRight.right == null) {
mostRight.right = cur;
cur = cur.left;
continue;
} else {
mostRight.right = null;
}
}
System.out.print(cur.value + " ");
cur = cur.right;
}
System.out.println();
}
public static void morrisPos(Node head) {
if (head == null) {
return;
}
Node cur = head;
Node mostRight = null;
while (cur != null) {
mostRight = cur.left;
if (mostRight != null) {
while (mostRight.right != null && mostRight.right != cur) {
mostRight = mostRight.right;
}
if (mostRight.right == null) {
mostRight.right = cur;
cur = cur.left;
continue;
} else {
mostRight.right = null;
printEdge(cur.left);
}
}
cur = cur.right;
}
printEdge(head);
System.out.println();
}
public static void printEdge(Node head) {
Node tail = reverseEdge(head);
Node cur = tail;
while (cur != null) {
System.out.print(cur.value + " ");
cur = cur.right;
}
reverseEdge(tail);
}
public static Node reverseEdge(Node from) {
Node pre = null;
Node next = null;
while (from != null) {
next = from.right;
from.right = pre;
pre = from;
from = next;
}
return pre;
}
// for test -- print tree
public static void printTree(Node head) {
System.out.println("Binary Tree:");
printInOrder(head, 0, "H", 17);
System.out.println();
}
public static void printInOrder(Node head, int height, String to, int len) {
if (head == null) {
return;
}
printInOrder(head.right, height + 1, "v", len);
String val = to + head.value + to;
int lenM = val.length();
int lenL = (len - lenM) / 2;
int lenR = len - lenM - lenL;
val = getSpace(lenL) + val + getSpace(lenR);
System.out.println(getSpace(height * len) + val);
printInOrder(head.left, height + 1, "^", len);
}
public static String getSpace(int num) {
String space = " ";
StringBuffer buf = new StringBuffer("");
for (int i = 0; i < num; i++) {
buf.append(space);
}
return buf.toString();
}
public static boolean isBST(Node head) {
if (head == null) {
return true;
}
Node cur = head;
Node mostRight = null;
Integer pre = null;
boolean ans = true;
while (cur != null) {
mostRight = cur.left;
if (mostRight != null) {
while (mostRight.right != null && mostRight.right != cur) {
mostRight = mostRight.right;
}
if (mostRight.right == null) {
mostRight.right = cur;
cur = cur.left;
continue;
} else {
mostRight.right = null;
}
}
if (pre != null && pre >= cur.value) {
ans = false;
}
pre = cur.value;
cur = cur.right;
}
return ans;
}
public static void main(String[] args) {
Node head = new Node(4);
head.left = new Node(2);
head.right = new Node(6);
head.left.left = new Node(1);
head.left.right = new Node(3);
head.right.left = new Node(5);
head.right.right = new Node(7);
printTree(head);
morrisIn(head);
morrisPre(head);
morrisPos(head);
printTree(head);
}
习题1 给定一棵二叉树的头节点head 求以head为头的树中，最小深度是多少？
public static class TreeNode {
public int val;
public TreeNode left;
public TreeNode right;
public TreeNode(int x) {
val = x;
}
}
// 下面的方法是一般解
public static int minDepth1(TreeNode head) {
if (head == null) {
return 0;
}
return p(head);
}
// 返回x为头的树，最小深度是多少
public static int p(TreeNode x) {
if (x.left == null && x.right == null) {
return 1;
}
// 左右子树起码有一个不为空
int leftH = Integer.MAX_VALUE;
if (x.left != null) {
leftH = p(x.left);
}
int rightH = Integer.MAX_VALUE;
if (x.right != null) {
rightH = p(x.right);
}
return 1 + Math.min(leftH, rightH);
}
// 下面的方法是morris遍历的解
public static int minDepth2(TreeNode head) {
if (head == null) {
return 0;
}
TreeNode cur = head;
TreeNode mostRight = null;
int curLevel = 0;
int minHeight = Integer.MAX_VALUE;
while (cur != null) {
mostRight = cur.left;
if (mostRight != null) {
int rightBoardSize = 1;
while (mostRight.right != null && mostRight.right != cur) {
rightBoardSize++;
mostRight = mostRight.right;
}
if (mostRight.right == null) { // 第一次到达
curLevel++;
mostRight.right = cur;
cur = cur.left;
continue;
} else { // 第二次到达
if (mostRight.left == null) {
minHeight = Math.min(minHeight, curLevel);
}
curLevel -= rightBoardSize;
mostRight.right = null;
}
} else { // 只有一次到达
curLevel++;
}
cur = cur.right;
}
int finalRight = 1;
cur = head;
while (cur.right != null) {
finalRight++;
cur = cur.right;
}
if (cur.left == null && cur.right == null) {
minHeight = Math.min(minHeight, finalRight);
}
return minHeight;
}