二叉樹的遍歷

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目錄

遞歸遍歷

非遞歸遍歷

Morris遍歷

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1. 遞歸遍歷

遞歸版遍歷只要當(dāng)前節(jié)點不為null，就可以三次回到當(dāng)前節(jié)點。

public static void preOrderRecur(Node head) {if (head == null) {return;}System.out.print(head.value + " ");preOrderRecur(head.left);preOrderRecur(head.right);}public static void inOrderRecur(Node head) {if (head == null) {return;}inOrderRecur(head.left);System.out.print(head.value + " ");inOrderRecur(head.right);}public static void posOrderRecur(Node head) {if (head == null) {return;}posOrderRecur(head.left);posOrderRecur(head.right);System.out.print(head.value + " ");}

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2. 非遞歸遍歷

其中后序遍歷打印順序為左右中，由先序遍歷打印順序為中左右，所以可以對先序遍歷改進為中右左（改變添加順序），添加到另外一個棧中，最后遍歷打印就是左右中順序了。

public static void preOrderUnRecur(Node head) {System.out.println("pre-order: ");while (head != null) {Stack<Node> stack = new Stack<>();stack.push(head);while (!stack.isEmpty()) {Node node = stack.pop();System.out.println(node.value + " ");if (node.right != null) {stack.push(node.right);}if (node.left != null) {stack.push(node.left);}}}}public static void inOrderUnRecur(Node head) {System.out.print("in-order: ");if (head != null) {Stack<Node> stack = new Stack<>();while (!stack.isEmpty() || head != null) {if (head != null) {stack.push(head.left);head = head.left;} else {head = stack.pop();System.out.println(head.value + " ");head = head.right;}}}System.out.println();}public static void posOrderUnRecur1(Node head) {System.out.print("pos-order: ");if (head != null) {Stack<Node> s1 = new Stack<>();Stack<Node> s2 = new Stack<>();s1.push(head);while (!s1.isEmpty()) {head = s1.pop();s2.push(head);if (head.left != null) {s1.push(head.left);}if (head.right != null) {s1.push(head.right);}while (!s2.isEmpty()) {System.out.print(s2.pop().value + " ");}}}System.out.println();}

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3. Morris遍歷

Morris遍歷法，能以時間復(fù)雜度O(N)，空間復(fù)雜度O(1)實現(xiàn)二叉樹的遍歷。

程序流程：
假設(shè)指針cur指向當(dāng)前節(jié)點，cur從頭結(jié)點開始。

如果cur無左孩子，則cur = cur.right；

如果cur有左孩子，則找到cur左子樹上最右的節(jié)點，記為mostRight，分為以下兩種情況：

若mostRight的right指針為null，則讓其指向cur，且cur = cur.left;

若mostRight的right指針指向cur，則讓其指回null，且cur = cur.right。

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假設(shè)二叉樹如下：

12 34 5 6 7

則Morris遍歷順序為：1,2,4,2,5,1,3,6,3,7

Morris 遍歷代碼實現(xiàn)：

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;}}cur = cur.right;}}

特點：

當(dāng)某個節(jié)點有左子樹，則會到達該節(jié)點兩次，如果沒有左子樹，只會到達一次。

當(dāng)?shù)诙位氐侥硞€節(jié)點時，它的左子樹已遍歷完。

實質(zhì)：
Morris遍歷是利用左子樹最右節(jié)點的指針指向null或指向自己來標(biāo)記是第一次來到該節(jié)點還是第二次來到該節(jié)點。

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Morris遍歷改先序遍歷

在以下兩種情況下打印節(jié)點:

當(dāng)節(jié)點沒有左子樹時，打印當(dāng)前節(jié)點；

當(dāng)節(jié)點有左子樹時并且第一次訪問時打印該節(jié)點，就是當(dāng)mosrRight的右指針為null時。

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) {mostRight.right = cur;System.out.print(cur.value + " ");cur = cur.left;continue;} else {mostRight.right = null;}} else {System.out.print(cur.value + " ");}cur = cur.right;}System.out.println();}

Morris遍歷改中序遍歷

在以下兩種情況下打印節(jié)點:

當(dāng)節(jié)點沒有左子樹時，說明第一次和第二次是重合在一起的，打印當(dāng)前節(jié)點即可。

當(dāng)節(jié)點有左子樹時，那么需要處理完左子樹再打印，即當(dāng)前節(jié)點準(zhǔn)備右移時打印。

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();}

Morris遍歷改后序遍歷

在以下兩種情況下打印節(jié)點:

只有當(dāng)?shù)竭_某個節(jié)點兩次時逆序打印該節(jié)點左子樹的右邊界;

在代碼的最后逆序打印整棵樹的右邊界，而逆序的過程就和單向鏈表的反轉(zhuǎn)過程類似。

public static void morrisPos(Node head) {if (head == null) {return;}Node cur1 = head;Node cur2 = null;while (cur1 != null) {cur2 = cur1.left;if (cur2 != null) {while (cur2.right != null && cur2.right != cur1) {cur2 = cur2.right;}if (cur2.right == null) {cur2.right = cur1;cur1 = cur1.left;continue;} else {cur2.right = null;printEdge(cur1.left);}}cur1 = cur1.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;}

總結(jié)

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