剑指offer 二叉树专题 刷题记录（1）

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode reConstructBinaryTree(int [] pre,int [] in) {
        TreeNode root=reConstructBinaryTree(pre,0,pre.length-1,in,0,in.length-1);
        return root;
    }
    public TreeNode reConstructBinaryTree(int[] pre,int preStart,int preEnd,int[] in,int inStart,int inEnd){
        if(preStart>preEnd||inStart>inEnd){
            return null;
        }
        TreeNode root=new TreeNode(pre[preStart]);
        for(int i=inStart;i<=inEnd;i++){
            if(pre[preStart]==in[i]){
                root.left=reConstructBinaryTree(pre,preStart+1,preStart+i-inStart,in,inStart,i-1);
                root.right=reConstructBinaryTree(pre,preStart+i-inStart+1,preEnd,in,i+1,inEnd);
                break;
            }
        }
        return root;
    }
}

public class Solution {
    public static boolean HasSubtree(TreeNode root1, TreeNode root2) {
    boolean result = false;
    //当Tree1和Tree2都不为零的时候，才进行比较。否则直接返回false
    if (root2 != null && root1 != null) {
      //如果找到了对应Tree2的根节点的点
      if(root1.val == root2.val){
        //以这个根节点为为起点判断是否包含Tree2
        result = doesTree1HaveTree2(root1,root2);
      }
      //如果找不到，那么就再去root的左儿子当作起点，去判断时候包含Tree2
      if (!result) {
        result = HasSubtree(root1.left,root2);
      }
      //如果还找不到，那么就再去root的右儿子当作起点，去判断时候包含Tree2
      if (!result) {
        result = HasSubtree(root1.right,root2);
         }
      }
        //返回结果
    return result;
  }
  public static boolean doesTree1HaveTree2(TreeNode node1, TreeNode node2) {
    //如果Tree2已经遍历完了都能对应的上，返回true
    if (node2 == null) {
      return true;
    }
    //如果Tree2还没有遍历完，Tree1却遍历完了。返回false
    if (node1 == null) {
      return false;
    }
    //如果其中有一个点没有对应上，返回false
      if (node1.val != node2.val) { 
        return false;
    }
      //如果根节点对应的上，那么就分别去子节点里面匹配
      return doesTree1HaveTree2(node1.left,node2.left) && doesTree1HaveTree2(node1.right,node2.right);
    }
}

import java.util.*;
public class Solution {
    public TreeNode Mirror (TreeNode pRoot) {
        if(pRoot==null)
            return pRoot;
        if(pRoot.left==null && pRoot.right==null)
            return pRoot;
        //处理根节点，交换左右节点
        TreeNode temp=pRoot.left;
        pRoot.left=pRoot.right;
        pRoot.right=temp;
        //处理左子树
        Mirror(pRoot.left);
        //处理右子树
        Mirror(pRoot.right);
        return pRoot;
    }
}

import java.util.ArrayList;
import java.util.Queue;
import java.util.LinkedList;
/**
public class TreeNode {
    int val = 0;
    TreeNode left = null;
    TreeNode right = null;
    public TreeNode(int val) {
        this.val = val;
    }
}
*/
public class Solution {
    public ArrayList<Integer> PrintFromTopToBottom(TreeNode root) {
        ArrayList<Integer> list=new ArrayList<Integer>();
        Queue<TreeNode> queue=new LinkedList<TreeNode>();
        if(root==null)
        {
            return list;
        }
        queue.add(root);
        while(!queue.isEmpty()){
            TreeNode temp=queue.poll();
            list.add(temp.val);
            if(temp.left!=null){
                queue.add(temp.left);
            }
            if(temp.right!=null){
                queue.add(temp.right);
            }
        }
        return list;
    }
}