1.输出二叉树的右视图

import java.util.*;


public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 求二叉树的右视图
     * @param preOrder int整型一维数组 先序遍历
     * @param inOrder int整型一维数组 中序遍历
     * @return int整型一维数组
     */
    public int[] solve (int[] preOrder, int[] inOrder) {
        // write code here
        List<Integer> list = new ArrayList<>();
        TreeNode root = createOneTree(preOrder, inOrder, 0, preOrder.length - 1, 0);
        layerOrder(root, list);
        int[] ans = new int[list.size()];
        for (int i = 0; i < list.size(); i++) {
            ans[i] = list.get(i);
        }
        return ans;
    }
    public TreeNode createOneTree(int[] preOrder, int[] inOrder, int left,
                                  int right, int preIndex) {
        if (left > right) return null;
        if (left == right) return new TreeNode(preOrder[preIndex]);
        int item = preOrder[preIndex];
        TreeNode node = new TreeNode(preOrder[preIndex]);
        for(int i=left;i<=right;i++){
            if(inOrder[i]==item){
                node.left=createOneTree(preOrder,inOrder,left,i-1,preIndex+1);
                node.right=createOneTree(preOrder,inOrder,i+1,right,preIndex+(i-left+1));
                break;
            }
        }
        return node;
    }
    public void layerOrder(TreeNode root, List<Integer> list) {
        if (root == null) return;
        List<TreeNode> nodeList = new ArrayList<>();
        nodeList.add(root);
        while (!nodeList.isEmpty()){
            int size=nodeList.size();
            list.add(nodeList.get(nodeList.size()-1).val);
            for(int i=0;i<size;i++){
                TreeNode remove = nodeList.remove(0);
                if(remove.left!=null) nodeList.add(remove.left);
                if(remove.right!=null) nodeList.add(remove.right);
            }
        }
    }
}
class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;
    public TreeNode(int val) {
        this.val = val;
    }
}

2.重建二叉树

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param preOrder int整型一维数组
     * @param vinOrder int整型一维数组
     * @return TreeNode类
     */
    public TreeNode reConstructBinaryTree (int[] preOrder, int[] vinOrder) {
        // write code here
        return createTree(preOrder,vinOrder,0, preOrder.length-1,0);
    }
    public TreeNode createTree(int[] preOrder, int[] vinOrder,int left,int right,int preIndex){
        if(left>right) return null;
        else if (left == right) {
            return new TreeNode(vinOrder[left]);
        }else{
            int item=preOrder[preIndex];
            int mid=0;
            for(int i=left;i<=right;i++){
                if(vinOrder[i]==item){
                    mid=i;
                }
            }
            //1 2 3 4 5
            //4 2 5 1 3
            TreeNode treeNode=new TreeNode(item);
            treeNode.left=createTree(preOrder,vinOrder,left,mid-1,preIndex+1);
            treeNode.right=createTree(preOrder,vinOrder,mid+1,right,preIndex+(mid-left+1));
            return treeNode;
        }
    }
}

3.在二叉树中找到两个节点的最近公共祖先

import java.util.*;

/*
 * public class TreeNode {
 *   int val = 0;
 *   TreeNode left = null;
 *   TreeNode right = null;
 *   public TreeNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param root TreeNode类
     * @param o1 int整型
     * @param o2 int整型
     * @return int整型
     */
    public int lowestCommonAncestor (TreeNode root, int o1, int o2) {
        // write code here
        if (root == null) {
            return 0;
        }
        if (root.val == o1) return root.val;
        if (root.val == o2) return root.val;
        int left = lowestCommonAncestor(root.left, o1, o2);
        int right = lowestCommonAncestor(root.right, o1, o2);
        if (left != 0 && right != 0) return root.val;
        if (left != 0 && right == 0) return left;
        if (left == 0 && right != 0) return right;
        return 0;
    }
}