LC.P105[从前序与中序遍历序列构造二叉树]

方法一：DFS+哈希表

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int[] preorder;
    int[] inorder;
    Map<Integer, Integer> map = new HashMap<>();

    public TreeNode buildTree(int[] preorder, int[] inorder) {
        this.preorder = preorder;
        this.inorder = inorder;
        int n = inorder.length;
        for (int i = 0; i < n; ++i) {
            map.put(inorder[i], i);
        }
        return build(0, n - 1, 0, n - 1);
    }

    private TreeNode build(int preorderLeft, int preorderRight, int inorderLeft, int inorderRight) {
        if (preorderLeft > preorderRight) return null;
        int inorderRoot = map.get(preorder[preorderLeft]); // 在中序遍历中定位根节点
        TreeNode root = new TreeNode(preorder[preorderLeft]);
        int leftSubTreeSize = inorderRoot - inorderLeft; // 左子树数目
        root.left = build(preorderLeft + 1, preorderLeft + leftSubTreeSize, inorderLeft, inorderRoot - 1);
        root.right = build(preorderLeft + leftSubTreeSize + 1, preorderRight, inorderRoot + 1, inorderRight);
        return root;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$