【LeetCode】 106. Construct Binary Tree from Inorder and Postorder Traversal 从中序与后序遍历序列构造二叉树(Medium)(JAVA)
题目地址: https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/
题目描述:
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
题目大意
根据一棵树的中序遍历与后序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。
解题方法
和上一题一样,只是前序遍历和后序遍历的属性不同:【LeetCode】 105. Construct Binary Tree from Preorder and Inorder Traversal 从前序与中序遍历序列构造二叉树(JAVA)
1、后序遍历的根节点在最后一个;中序遍历的根节点在中间,找到根节点,就能分为左右子树
2、迭代更新完整的树
/**
* 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[] inorder, int[] postorder) {
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < inorder.length; i++) {
map.put(inorder[i], i);
}
return bH(inorder, postorder, map, 0, inorder.length - 1, 0, postorder.length - 1);
}
public TreeNode bH(int[] inorder, int[] postorder, Map<Integer, Integer> map, int iStart, int iEnd, int pStart, int pEnd) {
if (iStart > iEnd || pStart > pEnd) return null;
TreeNode root = new TreeNode(postorder[pEnd]);
int mid = map.get(postorder[pEnd]);
root.left = bH(inorder, postorder, map, iStart, mid - 1, pStart, pStart + mid - iStart - 1);
root.right = bH(inorder, postorder, map, mid + 1, iEnd, pStart + mid - iStart, pEnd - 1);
return root;
}
}
执行用时 : 3 ms, 在所有 Java 提交中击败了 76.80% 的用户
内存消耗 : 40.4 MB, 在所有 Java 提交中击败了 61.90% 的用户