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Construct Binary Tree from Preorder and Inorder Traversal

Problem105

Given preorder and inorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

For example, given

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preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

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3
4
5
  3
/ \
9 20
/ \
15 7

key

题目是一个根据前序中序,生成二叉树的题目

前序遍历有个特点:根节点在前面,root -left-right

则遍历到3作为root,根据中序可以知道左子树是9,右子树是15 20 7

然后遍历9作为root,根据中序得到没有左子树,没有右子树

然后遍历20作为root,依次类推可以得到

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TreeNode root = new TreeNode(rootVal);
root.left = buildTree(pre, preStart+1, preStart+len, in, inStart, rootIndex-1);
root.right = buildTree(pre, preStart+len+1, preEnd, in, rootIndex+1, inEnd);

其中insort比较好理解,确定root后

左子树在inStart, rootIndex-1之间

右子树在rootIndex+1, inEnd之间

对于presort

int len = rootIndex - inStart;获得root的左子树长度(根据中序获取rootIndex)

左子树在preStart+1, preStart+len之间

右子树在preStart+len+1, preEnd之间

Solution

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/**
* 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[] preorder, int[] inorder) {
return buildTree(preorder, 0, preorder.length-1, inorder, 0, inorder.length-1);
}
public TreeNode buildTree(int[] pre, int preStart, int preEnd, int[] in, int inStart, int inEnd){
if(inStart > inEnd || preStart > preEnd)
return null;

int rootVal = pre[preStart];
int rootIndex = 0;
for(int i = inStart; i <= inEnd; i++){
if(in[i] == rootVal){
rootIndex = i;
break;
}
}

int len = rootIndex - inStart;
TreeNode root = new TreeNode(rootVal);
root.left = buildTree(pre, preStart+1, preStart+len, in, inStart, rootIndex-1);
root.right = buildTree(pre, preStart+len+1, preEnd, in, rootIndex+1, inEnd);

return root;
}
}

tip

参考于百度,在递归条件乱了