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_105_buildTree.java
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_105_buildTree.java
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package pp.arithmetic.leetcode;
import pp.arithmetic.Util;
import pp.arithmetic.model.TreeNode;
import java.util.Arrays;
/**
* Created by wangpeng on 2019-07-27.
* 105. 从前序与中序遍历序列构造二叉树
* <p>
* 根据一棵树的前序遍历与中序遍历构造二叉树。
* <p>
* 注意:
* 你可以假设树中没有重复的元素。
* <p>
* 例如,给出
* <p>
* 前序遍历 preorder = [3,9,20,15,7]
* 中序遍历 inorder = [9,3,15,20,7]
* 返回如下的二叉树:
* <p>
* 3
* / \
* 9 20
* / \
* 15 7
* <p>
* 来源:力扣(LeetCode)
* 链接:https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal
* 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
*/
public class _105_buildTree {
public static void main(String[] args) {
_105_buildTree buildTree = new _105_buildTree();
Util.printTree(buildTree.buildTree1(new int[]{3, 9, 20, 15, 7}, new int[]{9, 3, 15, 20, 7}));
//优化方案
Util.printTree(buildTree.buildTree(new int[]{1, 2, 3}, new int[]{3, 2, 1}));
Util.printTree(buildTree.buildTree(new int[]{3, 9, 20, 15, 7}, new int[]{9, 3, 15, 20, 7}));
}
/**
* 解题思路:
* 1.确定根节点,前序的首位就是根节点
* 2.确定根节点的左子树和右子树,找到中序根节点之前的就是左子树,之后的就是右子树
* 3.将左子树和右子树的前序和中序传递到buildTree中,递归1-2步骤
* <p>
* 执行耗时
* 执行用时 :48 ms, 在所有 Java 提交中击败了9.05%的用户
* 内存消耗 :77.6 MB, 在所有 Java 提交中击败了5.07%的用户
* <p>
* 优化:提交时间耗时有点长,感觉问题出在数组拷贝下,将数组拷贝改为传递下标试试
* {@link _105_buildTree#buildTree(int[], int[])}
*
* @param preorder
* @param inorder
* @return
*/
public TreeNode buildTree1(int[] preorder, int[] inorder) {
if (preorder.length == 0) return null;
//1
TreeNode root = new TreeNode(preorder[0]);
if (preorder.length == 1) return root;
//2
int leftIndex;
for (leftIndex = 0; leftIndex < inorder.length; leftIndex++) {
if (inorder[leftIndex] == preorder[0]) {
break;
}
}
//3
TreeNode leftNode = buildTree1(Arrays.copyOfRange(preorder, 1, leftIndex + 1), Arrays.copyOfRange(inorder, 0, leftIndex));
TreeNode rightNode = buildTree1(Arrays.copyOfRange(preorder, leftIndex + 1, preorder.length), Arrays.copyOfRange(inorder, leftIndex + 1, inorder.length));
root.left = leftNode;
root.right = rightNode;
return root;
}
/**
* 优化方案:避免数组的拷贝
*
* 执行用时 :3 ms, 在所有 Java 提交中击败了99.12%的用户
* 内存消耗 :37.5 MB, 在所有 Java 提交中击败了70.21%的用户
*
* @param preorder
* @param inorder
* @return
*/
public TreeNode buildTree(int[] preorder, int[] inorder) {
return buildTree(preorder.length, preorder, 0, inorder, 0);
}
// 从前序和中序构造二叉树,前序和中序是大数组中的一段[start, start + count)
private TreeNode buildTree(int count, int[] preOrder, int preStart, int[] inOrder, int inStart) {
if (count <= 0) return null;
int rootValue = preOrder[preStart];
TreeNode root = new TreeNode(rootValue);
// 从inorder中找到root值,(inorder)左边就是左子树,(inorder)右边就是右子树
// 然后在preorder中,数出与inorder中相同的个数即可
int pos = inStart + count - 1;
for (; pos >= inStart; --pos) {
if (inOrder[pos] == rootValue) {
break;
}
}
int leftCount = pos - inStart;
int rightCount = inStart + count - pos - 1;
if (leftCount > 0) {
int leftInStart = inStart;
int leftPreStart = preStart + 1;
root.left = buildTree(leftCount, preOrder, leftPreStart, inOrder, leftInStart);
}
if (rightCount > 0) {
int rightInStart = pos + 1;
int rightPreStart = preStart + 1 + leftCount;
root.right = buildTree(rightCount, preOrder, rightPreStart, inOrder, rightInStart);
}
return root;
}
}