复原二叉树

Posted 2020-10-27 midiyu

1.由（preorder+inorder）复原

/*
 * preorder:先序遍历数组
 * ps,pe:先序数组的首元素和末元素的索引
 * inod：中序遍历数组
 * is,ie:中序数组的首元素和末元素的索引 
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        return buildBT(preorder,0,preorder.length-1,inorder,0,inorder.length-1);
    }
    
    private TreeNode buildBT(int[] preorder,int ps,int pe,int[] inorder,int is,int ie) {
        if(pe<ps)
            return null;
        TreeNode root=new TreeNode(preorder[ps]);
        if(pe==ps)
            return root;
        int middle=0;
        for(int i=is;i<=ie;i++)
            if(inorder[i]==preorder[ps]) {
                middle=i;
                break;
            }
        int len=middle-is;
        root.left=buildBT(preorder,ps+1,ps+len,inorder,is,middle-1);
        root.right=buildBT(preorder,ps+len+1,pe,inorder,middle+1,ie);
        return root;
    }
}

 

 

2.由（inorder+postorder）复原

/**
 * 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) {
        return buildBT(inorder, 0, inorder.length-1, postorder, 0, postorder.length-1);
    }
    
    private TreeNode buildBT(int[] inorder,int is,int ie,int[] postorder,int ps,int pe) {
        if(is>ie)
            return null;
        TreeNode root=new TreeNode(postorder[pe]);
        if(pe==ps)
            return root;
        int middle=0;
        for(int i=is;i<=ie;i++)
            if(inorder[i]==postorder[pe]) {
                middle=i;
                break;
            }
        int len=middle-is;
        root.left=buildBT(inorder,is,middle-1,postorder,ps,ps+len-1);
        root.right=buildBT(inorder,middle+1,ie,postorder,ps+len,pe-1);
        return root;
    }
}