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Construct Binary Tree from Inorder and Postorder Traversal

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.

分析

代码

// Construct Binary Tree from Inorder and Postorder Traversal
// 递归,时间复杂度O(n),空间复杂度O(\logn)
public class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {
return buildTree(inorder, 0, inorder.length,
postorder, 0, postorder.length);
}

TreeNode buildTree(int[] inorder, int begin1, int end1,
int[] postorder, int begin2, int end2) {
if (begin1 ==end1) return null;
if (begin2 ==end2) return null;

int val = postorder[end2 - 1];
TreeNode root = new TreeNode(val);

int in_root_pos = find(inorder, begin1, end1, val);
int left_size = in_root_pos - begin1;
int post_left_last = begin2 + left_size;

root.left = buildTree(inorder, begin1, in_root_pos,
postorder, begin2, post_left_last);
root.right = buildTree(inorder, in_root_pos + 1, end1,
postorder, post_left_last, end2 - 1);

return root;
}
private static int find(int[] array, int begin, int end, int val) {
for (int i = begin; i < end; ++i) {
if (array[i] == val) return i;
}
return -1;
}
}

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