从前序遍历与中序遍历构造二叉树

105. 从前序与中序遍历序列构造二叉树

难度: 中等

根据一棵树的前序遍历与中序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

例如，给出

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：

  3
 / \
9  20
  /  \
 15   7

根据前序遍历找根，（使用哈希表记录中序遍历每个元素的index）根据根找左子树的元素和右子树的元素

难度在于处理边界条件

/**
 * 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) {

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

        TreeNode node = buildTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1, map);

        return node;
    }

    public TreeNode buildTree(int[] preorder, int pStart, int pEnd, int[] inorder, int iStart, int iEnd, Map<Integer, Integer> val2Index) {

        if (pStart > pEnd) {
            return null;
        }

        TreeNode root = new TreeNode(preorder[pStart]);
        if (pStart == pEnd) {
            return root;
        }

        int rootIndex = val2Index.get(root.val);
        int leftNum = rootIndex - iStart;
        int rightNum = iEnd - rootIndex;

        TreeNode left = buildTree(preorder, pStart + 1, pStart + leftNum, inorder, iStart, rootIndex - 1, val2Index);
        TreeNode right = buildTree(preorder, pEnd - rightNum + 1, pEnd, inorder, rootIndex + 1, iEnd, val2Index);

        root.left = left;
        root.right = right;
        return root;
    }
}

赏

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