105. Construct Binary Tree from Preorder and Inorder Traversal - jiejackyzhang/leetcode-note GitHub Wiki

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

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

解题思路为：preorder的第一个元素即为root，在inorder中找出root的位置，它的左右即为左右子树。 剩下的过程可通过recursion完成。

/**
 * 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(inorder == null || preorder == null || inorder.length != preorder.length) return null;
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for(int i = 0; i < inorder.length; i++) {
            map.put(inorder[i], i);
        }
        return buildTreeHelper(preorder, 0, preorder.length-1, inorder, 0, inorder.length-1, map);
    }
    
    private TreeNode buildTreeHelper(int[] preorder, int ps, int pe, int[] inorder, int is, int ie, Map<Integer, Integer> map) {
        if(ps > pe || is > ie) return null;
        TreeNode root = new TreeNode(preorder[ps]);
        int ri = map.get(preorder[ps]);
        root.left = buildTreeHelper(preorder, ps+1, ps+ri-is, inorder, is, ri-1, map);
        root.right = buildTreeHelper(preorder, ps+ri-is+1, pe, inorder, ri+1, ie, map);
        return root;
    }
}