剑指offer_04:重建二叉树

标签：init   中序遍历   限制   index   节点   else   ack   public   dex   

输入某二叉树的前序遍历和中序遍历的结果，请重建该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。
例如，给出
前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]

限制：
0 <= 节点个数 <= 5000

1、递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder.length==0) return null;
        Map<Integer,Integer> map=new HashMap<>();
        for(int i=0;i<inorder.length;i++){
            map.put(inorder[i],i);
        }
        TreeNode root=build(preorder,0,preorder.length-1,inorder,0,inorder.length-1,map);
        return root;
    }
    public TreeNode build(int[] preorder,int ps,int pe,int[] inorder,int is,int ie,Map<Integer,Integer> map){
        if(ps>pe){
            return null;
        }
        int rootval=preorder[ps];
        TreeNode root=new TreeNode(rootval);
        if(ps==pe){
            return root;
        }else{
            int index=map.get(rootval);
            int lefts=index-is;
            int rights=ie-index;
            root.left=build(preorder,ps+1,ps+lefts,inorder,is,index-1,map);
            root.right=build(preorder,ps+lefts+1,pe,inorder,index+1,ie,map);
            return root;
        }
    }
}

2、迭代，使用栈

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder.length==0) return null;
        TreeNode root=new TreeNode(preorder[0]);
        Stack<TreeNode> stack=new Stack<>();
        int len=preorder.length;
        stack.push(root);
        int index=0;
        for(int i=1;i<len;i++){
            int pre=preorder[i];
            TreeNode node=stack.peek();
            if(node.val!=inorder[index]){
                node.left=new TreeNode(pre);
                stack.push(node.left);
            }else{
                while(!stack.isEmpty()&&stack.peek().val==inorder[index]){
                    node=stack.pop();
                    index++;
                }
                node.right=new TreeNode(pre);
                stack.push(node.right);
            }
        }
        return root;
    }
}

剑指offer_04:重建二叉树

标签：init   中序遍历   限制   index   节点   else   ack   public   dex   

原文地址：https://www.cnblogs.com/xyz-1024/p/13956416.html