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;    }}