力扣labuladong——一刷day74

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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);
    }
}