【重点】【二叉树】889. 根据前序和后序遍历构造二叉树

题目

法1：递归

Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def constructFromPrePost(self, preorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
        if len(preorder) == 0:
            return None
        elif len(preorder) == 1:
            return TreeNode(preorder[0])
        root_val = preorder[0]
        left_root_val = preorder[1]
        left_size = postorder.index(left_root_val) + 1
        left = self.constructFromPrePost(preorder[1: 1+left_size], postorder[:left_size])
        right = self.constructFromPrePost(preorder[1+left_size:], postorder[left_size: -1])

        return TreeNode(root_val, left, right)

Java

class Solution {
    public TreeNode constructFromPrePost(int[] preorder, int[] postorder) {
        return build(preorder, 0, preorder.length - 1, postorder, 0, postorder.length - 1);
    }

    public TreeNode build(int[] preorder, int preStart, int preEnd, int[] postorder, int postStart, int postEnd) {
        if (preStart > preEnd) {
            return null;
        }
        if (preStart == preEnd) {
            return new TreeNode(preorder[preStart]);
        }
        TreeNode root = new TreeNode(preorder[preStart]);
        int lastLeftInx = -1;
        for (int i = postStart; i <= postEnd; ++i) {
            if (postorder[i] == preorder[preStart + 1]) {
                lastLeftInx = i;
                break;
            }
        }
        int leftSize = lastLeftInx - postStart + 1;
        root.left = build(preorder, preStart + 1, preStart + leftSize, postorder, postStart, lastLeftInx);
        root.right = build(preorder, preStart + leftSize + 1, preEnd, postorder, lastLeftInx + 1, postEnd - 1);
        return root;
    }

}