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Construct Binary Tree from Preorder and Inorder Traversal

描述

Given preorder and inorder 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 Preorder and Inorder Traversal
// 递归,时间复杂度O(n),空间复杂度O(\logn)
public class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
return buildTree(preorder, 0, preorder.length,
inorder, 0, inorder.length);
}

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

TreeNode root = new TreeNode(preorder[begin1]);

int inRootPos = find(inorder, begin2, end2, preorder[begin1]);
int leftSize = inRootPos - begin2;

root.left = buildTree(preorder, begin1 + 1, begin1 + leftSize + 1,
inorder, begin2, begin2 + leftSize);
root.right = buildTree(preorder, begin1 + leftSize + 1, end1,
inorder, inRootPos + 1, end2);

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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