889. Construct Binary Tree from Preorder and Postorder Traversal

题目

Return any binary tree that matches the given preorder and postorder traversals.

Values in the traversals pre and post are distinct positive integers.

Example 1:

Input: pre = [1,2,4,5,3,6,7], post = [4,5,2,6,7,3,1]
Output: [1,2,3,4,5,6,7]

Note:

• 1 <= pre.length == post.length <= 30
• pre[] and post[] are both permutations of 1, 2, ..., pre.length.
• It is guaranteed an answer exists. If there exists multiple answers, you can return any of them.

讲解

先序遍历和后序遍历无法确定一颗二叉树，所以这里有可能返回多种不同的树，但只要是正确的就行。

解这道题的关键在于分割左右子树，先序遍历中紧跟根节点后的也有可能不是左子树，但如果是这样的话，后续遍历中也不会有左子树，那么右子树的根节点就会在两种遍历中都挨着总根节点。所以思路就出来了，我们在先序遍历中拿到根节点后面的那个节点，然后在后序遍历中找到这个节点，这个点就在后序遍历中分割出了左右子树。

递归之。

Java代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode constructFromPrePost(int[] pre, int[] post) {
        return constructFromPrePost(pre, 0, pre.length-1, post, 0, post.length-1);
    }
    
    private TreeNode constructFromPrePost(int[] pre, int preLeft, int preRight, int[] post, int postLeft, int postRight){
        if(preLeft>preRight || postLeft>postRight){
            return null;
        }else if(preLeft==preRight){
            return new TreeNode(pre[preLeft]);
        }
        TreeNode root = new TreeNode(pre[preLeft]);
        for(int i=postRight-1;i>=postLeft;i--){
            if(post[i]==pre[preLeft+1]){
                root.left = constructFromPrePost(pre, preLeft+1, preLeft+1+i-postLeft, post, postLeft, i);
                root.right = constructFromPrePost(pre, preLeft+2+i-postLeft, preRight, post, i+1, postRight-1);
            }
        }
        return root;
    }
}