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1 后繼節(jié)點

1.1 解題思路

二叉樹節(jié)點結(jié)構(gòu)定義如下：

public static class Node {
public int cal;
public Node left;
public Node right;
public Node parent;
}
給你二叉樹中的某個節(jié)點，返回該節(jié)點的后繼節(jié)點

后繼節(jié)點就是二叉樹中序遍歷，這個節(jié)點的下一個節(jié)點

思路

1. 如果該節(jié)點有右子樹，那么后繼節(jié)點就是右樹的最左節(jié)點
2. 如果該節(jié)點沒有右子樹
   1. 如果該節(jié)點是父節(jié)點的左節(jié)點，此時父節(jié)點就是后繼節(jié)點
   2. 如果該節(jié)點是父節(jié)點的右節(jié)點，向上尋找
      1. 這個節(jié)點在頂部節(jié)點的右樹上，此時返回的是空
      2. 如果這個節(jié)點在某個節(jié)點的左數(shù)上，返回此時的節(jié)點

1.2 代碼實現(xiàn)

public class SuccessorNode {public static class Node {public int val;public Node left;public Node right;public Node parent;public Node(int val) {this.val = val;}}public static Node getSuccessorNode(Node node) {if (node == null) {return node;}if (node.right != null) {return getLeftMost(node.right);}else{Node cur = node;Node parent = node.parent;while (parent != null && parent.left != node) {cur = parent;parent = cur.parent;}return parent;}}private static Node getLeftMost(Node node) {Node cur = node;if (cur.left != null) {cur = cur.left;}return cur;}// 測試public static void main(String[] args) {Node n1 = new Node(1);Node n2 = new Node(2);Node n3 = new Node(3);Node n4 = new Node(4);Node n5 = new Node(5);Node n6 = new Node(6);Node n7 = new Node(7);n1.left = n2;n1.right = n3;n2.left = n4;n2.right = n5;n2.parent = n1;n3.left = n6;n3.right = n7;n3.parent = n1;n4.parent = n2;n5.parent = n2;n6.parent = n3;n7.parent = n3;System.out.println(getSuccessorNode(n6).val);}
}

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💎總結(jié)

本文中若是有出現(xiàn)的錯誤請在評論區(qū)或者私信指出，我再進行改正優(yōu)化，如果文章對你有所幫助，請給博主一個寶貴的三連，感謝大家😘！！！

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