力扣labuladong——一刷day74

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文章目录

前言
一、力扣1305. 两棵二叉搜索树中的所有元素
二、力扣872. 叶子相似的树

前言

二叉树的构造算法通用思路很简单，无非就是构造根节点，然后递归构造左右子树，最后把它们接起来，关键在于如何找到根节点和左右子树的节点，不同的序列化方法，找根节点的方式也不同

一、力扣1305. 两棵二叉搜索树中的所有元素

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {List<Integer> l1 = new ArrayList<>();List<Integer> l2 = new ArrayList<>();public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {fun(root1, l1);fun(root2, l2);List<Integer> res = new ArrayList<>();int index1 = 0, index2 = 0;while(index1 < l1.size() && index2 < l2.size()){if(l1.get(index1) < l2.get(index2)){res.add(l1.get(index1));index1 ++;}else{res.add(l2.get(index2));index2 ++;}}while(index1 < l1.size()){res.add(l1.get(index1));index1 ++;}while(index2 < l2.size()){res.add(l2.get(index2));index2 ++;}return res;}public void fun(TreeNode root, List<Integer> l){if(root == null){return;}fun(root.left, l);l.add(root.val);fun(root.right, l);}
}

二、力扣872. 叶子相似的树

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public boolean leafSimilar(TreeNode root1, TreeNode root2) {List<Integer> l1 = new ArrayList<>();List<Integer> l2 = new ArrayList<>();fun(root1, l1);fun(root2, l2);int index = 0;while(index < l1.size() && index < l2.size()){if(l1.get(index) != l2.get(index)){return false;}index ++;}if(index < l1.size() || index < l2.size()){return false;}return true;}public void fun(TreeNode root, List<Integer> l){if(root == null){return ;}fun(root.left,l);if(root.left == null && root.right == null){l.add(root.val);}fun(root.right, l);}
}