數據結構與算法之二叉樹的先序遍歷，中序遍歷，后移遍歷

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目錄

實現二叉樹的先序，中序，后序遍歷，包括遞歸方式和非遞歸方式

在二叉樹中找到一個節點的后繼節點

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1. 實現二叉樹的先序，中序，后序遍歷，包括遞歸方式和非遞歸方式

1. 先序遍歷，中序遍歷，后序遍歷遞歸版

//先序遍歷遞歸版public static void preOrderRecur(Node head) {if (head == null) {return;}System.out.print(head.value + " ");preOrderRecur(head.left);preOrderRecur(head.right);}//中序遍歷遞歸版public static void inOrderRecur(Node head) {if (head == null) {return;}inOrderRecur(head.left);System.out.print(head.value + " ");inOrderRecur(head.right);}//后序遍歷遞歸版public static void posOrderRecur(Node head) {if (head == null) {return;}posOrderRecur(head.left);posOrderRecur(head.right);System.out.print(head.value + " ");}

2. 先序遍歷，中序遍歷，后序遍歷非遞歸版

//先序遍歷非遞歸版public static void preOrderUnRecur(Node head) {System.out.print("pre-order: ");if (head != null) {Stack<Node> stack = new Stack<Node>();stack.add(head);while (!stack.isEmpty()) {head = stack.pop();System.out.print(head.value + " ");if (head.right != null) {stack.push(head.right);}if (head.left != null) {stack.push(head.left);}}}System.out.println();}//中序遍歷非遞歸版public static void inOrderUnRecur(Node head) {System.out.print("in-order: ");if (head != null) {Stack<Node> stack = new Stack<Node>();while (!stack.isEmpty() || head != null) {if (head != null) {stack.push(head);head = head.left;} else {head = stack.pop();System.out.print(head.value + " ");head = head.right;}}}System.out.println();}//后序遍歷非遞歸版1public static void posOrderUnRecur1(Node head) {System.out.print("pos-order: ");if (head != null) {Stack<Node> s1 = new Stack<Node>();Stack<Node> s2 = new Stack<Node>();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();}//后序遍歷非遞歸版2public static void posOrderUnRecur2(Node h) {System.out.print("pos-order: ");if (h != null) {Stack<Node> stack = new Stack<Node>();stack.push(h);Node c = null;while (!stack.isEmpty()) {c = stack.peek();if (c.left != null && h != c.left && h != c.right) {stack.push(c.left);} else if (c.right != null && h != c.right) {stack.push(c.right);} else {System.out.print(stack.pop().value + " ");h = c;}}}System.out.println();}

3. 先序，中序，后序遍歷非遞歸解析

先序遍歷
1. 如果節點不為null，則參加棧來存儲節點。
2. 先將頭節點添加到棧。
3. 當棧不為null時，從棧中彈出一個頭節點
4. 由棧特性先進后出和先序遍歷中左右知，先添加右節點，再添加左節點，這樣彈出的時候就是先左節點，然后右節點。

中序遍歷

如果head不為null，則創建棧結構進行存儲。

當棧或者head不為null時，如果head不為null，則將它的左孩子全部壓入棧。

當head==null時，則彈出棧頂節點，將head指向它的右孩子。

總結就是：當前節點為null，從棧中拿一個節點，打印，當前節點向右移。當前節點不為null，將它壓入棧，當前節點向左移；

后序遍歷

由先序遍歷為中左右，后序遍歷為左右中，我們可以通過兩個棧，使進入棧1的順序為中右左，然后將棧1元素放入棧2，就實現了左右中的結構。

如果head不為null，創建兩個棧s1，s2,。

將head先壓入s1。

當棧s1不為null時，棧s1彈出一個元素，將彈出的元素加入棧s2（即先序遍歷打印的位置）

判斷head左節點是否為null，不為null就加入s1，判斷head右節點是否為null，不為null就加入s1.（先加左，后加右，出來就是先右，后左）

打印出s2中的元素即可。

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2. 在二叉樹中找到一個節點的后繼節點

題目描述

思路：

node節點如果有右孩子，則右孩子的最左節點就是后繼節點。如1的后繼節點是右孩子3的最左節點6

node節點如果沒有右孩子，則往上找，找到當前節點是父節點的左孩子就停，那個父節點就是后繼節點。如5的后繼節點是1的左孩子是2，所以5的后繼節點是1

備注：前驅的找法：

node節點如果有左孩子，那么左孩子最右節點就是前驅節點，如1的前驅節點是左孩子2的最右節點5

如果node節點沒有左孩子，則往上找，找到當前節點是父節點的右孩子就停，那個父節點就是前驅節點，如6的前驅節點是1的右孩子3，所以6的前驅節點是1

代碼實現

public class Code_SuccessorNode {public static class Node {public int value;public Node left;public Node right;public Node parent;public Node(int data) {this.value = data;}}public static Node getSuccessorNode(Node node) {if (node == null) {return node;}if (node.right != null) {return getLeftMost(node.right);} else {Node parent = node.parent;while (parent != null && parent.left != node) {node = parent;parent = node.parent;}return parent;}}public static Node getLeftMost(Node node) {if (node == null) {return node;}while (node.left != null) {node = node.left;}return node;}public static void main(String[] args) {Node head = new Node(6);head.parent = null;head.left = new Node(3);head.left.parent = head;head.left.left = new Node(1);head.left.left.parent = head.left;head.left.left.right = new Node(2);head.left.left.right.parent = head.left.left;head.left.right = new Node(4);head.left.right.parent = head.left;head.left.right.right = new Node(5);head.left.right.right.parent = head.left.right;head.right = new Node(9);head.right.parent = head;head.right.left = new Node(8);head.right.left.parent = head.right;head.right.left.left = new Node(7);head.right.left.left.parent = head.right.left;head.right.right = new Node(10);head.right.right.parent = head.right;Node test = head.left.left;System.out.println(test.value + " next: " + getSuccessorNode(test).value);test = head.left.left.right;System.out.println(test.value + " next: " + getSuccessorNode(test).value);test = head.left;System.out.println(test.value + " next: " + getSuccessorNode(test).value);test = head.left.right;System.out.println(test.value + " next: " + getSuccessorNode(test).value);test = head.left.right.right;System.out.println(test.value + " next: " + getSuccessorNode(test).value);test = head;System.out.println(test.value + " next: " + getSuccessorNode(test).value);test = head.right.left.left;System.out.println(test.value + " next: " + getSuccessorNode(test).value);test = head.right.left;System.out.println(test.value + " next: " + getSuccessorNode(test).value);test = head.right;System.out.println(test.value + " next: " + getSuccessorNode(test).value);test = head.right.right; // 10's next is nullSystem.out.println(test.value + " next: " + getSuccessorNode(test));}}

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