如何找二叉树的后继结点

二叉树的后继结点的定义：为该二叉树的中序遍历后，该节点后面一个结点即使该节点的后继结点。

如图：中序遍历的结果是，4251637，所以 4的后继结点是2 ，2的后继结点是5

package class_04;

public class Code_03_SuccessorNode {

	public static class Node {
		public int value;
		public Node left;
		public Node right;
		public Node parent;   // 添加一个后继结点

		public Node(int data) {
			this.value = data;
		}
	}

	public static Node getSuccessorNode(Node node) {  // 找后继结点
		if (node == null) {
			return node;
		}
		if (node.right != null) {       // 如果有右子树，找到右子树上最左的结点
			return getLeftMost(node.right);
		} else {                    
			Node parent = node.parent;  
			while (parent != null && parent.left != node) {  // 往上找，直到“当前结点的父节点的左结点” 等于“当前结点”
				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 null
		System.out.println(test.value + " next: " + getSuccessorNode(test));
	}

}

左神的福利打印二叉树。

package class_04;

public class Code_03_SuccessorNode {

	public static class Node {
		public int value;
		public Node left;
		public Node right;
		public Node parent;   // 添加一个后继结点

		public Node(int data) {
			this.value = data;
		}
	}

	public static Node getSuccessorNode(Node node) {  // 找后继结点
		if (node == null) {
			return node;
		}
		if (node.right != null) {       // 如果有右子树，找到右子树上最左的结点
			return getLeftMost(node.right);
		} else {                    
			Node parent = node.parent;  
			while (parent != null && parent.left != node) {  // 往上找，直到“当前结点的父节点的左结点” 等于“当前结点”
				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 null
		System.out.println(test.value + " next: " + getSuccessorNode(test));
	}

}