二叉树-通过中序遍历和后序遍历||前序遍历和中序遍历构造二叉树java实现

思路：中序遍历顺序为左子树-根-右子树， 后序遍历顺序为 左子树-右子树-根，故通过后序遍历可以确定树的根节点，再通过根节点在中序遍历中确定左子树和右子树，递归实现以上过程完成树的构造。

图源：https://blog.csdn.net/qq_41855420/article/details/87717203

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder==null||postorder==null)
            return null;
        return buildtree(inorder,0,inorder.length-1,postorder,0,postorder.length-1);
    }
    public TreeNode buildtree(int[] inorder,int instart,int inend,int[] postorder,int postart,int poend){
        if(instart>inend)
            return null;
        int value=postorder[poend];
        int middle=instart;
        int pomiddle=postart;
        TreeNode root=new TreeNode(value);
        for(int i=instart;i<=inend;i++){
            if(inorder[i]==value)
                break;
            middle++;
            pomiddle++;
        }
        root.left=buildtree(inorder,instart,middle-1,postorder,postart,pomiddle-1);
        root.right=buildtree(inorder,middle+1,inend,postorder,pomiddle,poend-1);
        return root;
    }
}

根据前序和中序遍历同理，根节点在前序的首位：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder==null||inorder==null)
            return null;
        return buildtree(preorder,0,preorder.length-1,inorder,0,inorder.length-1);
    }
    public TreeNode buildtree(int[] preorder,int prestart,int preend,int[] inorder,int instart,int inend){
        if(prestart>preend||instart>inend)
            return null;
        int value=preorder[prestart];
        TreeNode root=new TreeNode(value);
        int middle=prestart;
        int inmiddle=instart;
        for(int i=instart;i<=inend;i++){
            if(inorder[i]==value)
                break;
            middle++;
            inmiddle++;
        }
        root.left=buildtree(preorder,prestart+1,middle,inorder,instart,inmiddle-1);
        root.right=buildtree(preorder,middle+1,preend,inorder,inmiddle+1,inend);
        return root;
    }
}