2021-05-17 13:03:27 +08:00
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//在二叉树中,根节点位于深度 0 处,每个深度为 k 的节点的子节点位于深度 k+1 处。
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//
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// 如果二叉树的两个节点深度相同,但 父节点不同 ,则它们是一对堂兄弟节点。
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//
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// 我们给出了具有唯一值的二叉树的根节点 root ,以及树中两个不同节点的值 x 和 y 。
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//
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// 只有与值 x 和 y 对应的节点是堂兄弟节点时,才返回 true 。否则,返回 false。
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//
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//
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//
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// 示例 1:
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//
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//
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//
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//输入:root = [1,2,3,4], x = 4, y = 3
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//输出:false
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//
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//
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// 示例 2:
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//
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//
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//
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//输入:root = [1,2,3,null,4,null,5], x = 5, y = 4
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//输出:true
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//
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//
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// 示例 3:
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//
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//
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//
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//
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//输入:root = [1,2,3,null,4], x = 2, y = 3
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//输出:false
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//
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//
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//
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// 提示:
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//
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//
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// 二叉树的节点数介于 2 到 100 之间。
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// 每个节点的值都是唯一的、范围为 1 到 100 的整数。
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//
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//
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//
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// Related Topics 树 广度优先搜索
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// 👍 172 👎 0
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package leetcode.editor.cn;
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import com.code.leet.entiy.TreeNode;
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//993:二叉树的堂兄弟节点
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public class CousinsInBinaryTree {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new CousinsInBinaryTree().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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/**
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* Definition for a binary tree node.
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* public class TreeNode {
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* int val;
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* TreeNode left;
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* TreeNode right;
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* TreeNode() {}
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* TreeNode(int val) { this.val = val; }
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* TreeNode(int val, TreeNode left, TreeNode right) {
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* this.val = val;
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* this.left = left;
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* this.right = right;
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* }
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* }
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*/
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class Solution {
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public boolean isCousins(TreeNode root, int x, int y) {
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2021-05-17 13:06:23 +08:00
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int[][] paraent = getParent(root, x, y, new int[2][2], 1,-1);
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2021-05-17 13:03:27 +08:00
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return paraent[0][0] != paraent[1][0] && paraent[0][1] == paraent[1][1];
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}
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2021-05-17 13:06:23 +08:00
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private int[][] getParent(TreeNode root, int x, int y, int[][] paraent, int deep,int before) {
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2021-05-17 13:03:27 +08:00
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if (paraent[1][1] > 0) {
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return paraent;
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}
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if (root.val == x || root.val == y) {
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x = root.val == x ? 0 : x;
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y = root.val == y ? 0 : y;
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if (paraent[0][1] > 0) {
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2021-05-17 13:06:23 +08:00
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paraent[1][0] = before;
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2021-05-17 13:03:27 +08:00
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paraent[1][1] = deep;
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} else {
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2021-05-17 13:06:23 +08:00
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paraent[0][0] = before;
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2021-05-17 13:03:27 +08:00
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paraent[0][1] = deep;
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}
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}
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if (root.left != null) {
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2021-05-17 13:06:23 +08:00
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paraent = getParent(root.left, x, y, paraent, deep + 1,root.val);
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2021-05-17 13:03:27 +08:00
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}
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if (paraent[1][1] > 0) {
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return paraent;
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}
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if (root.right != null) {
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2021-05-17 13:06:23 +08:00
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paraent = getParent(root.right, x, y, paraent, deep + 1,root.val);
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2021-05-17 13:03:27 +08:00
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}
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return paraent;
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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