LeetCode:Construct Binary Tree from Preorder and Inorder Traversal

Construct Binary Tree from Preorder and Inorder Traversal

Total Accepted: 66121 Total
Submissions: 227515 Difficulty: Medium

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

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

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 (M) Construct Binary Tree from Inorder and Postorder
Traversal

思路：

根据“前序”数组的第一个结点，在“中序”数组中找到“根”结点index；根据index把“前序”分成index的左右子树。

java code:

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {

if(preorder == null || inorder == null || preorder.length != inorder.length) return null;

HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
for(int i=0;i<inorder.length;i++)
map.put(inorder[i], i);

return buildTree(preorder, 0, preorder.length-1, inorder, 0, inorder.length-1, map);
}

// 自定义函数
private TreeNode buildTree(int[] preorder, int s1, int e1, int[] inorder, int s2, int e2, Map<Integer, Integer> map) {

if(s1 > e1 || s2 > e2) return null;

TreeNode root = new TreeNode(preorder[s1]);
int rootIndex = map.get(preorder[s1]); // 在前序数组中找到根结点

root.left = buildTree(preorder, s1+1, s1+rootIndex-s2, inorder, s2, rootIndex-1, map);
root.right = buildTree(preorder, s1+rootIndex-s2+1, e1, inorder, rootIndex+1, e2, map);
return root;
}
}