Tree-----重建树（根据中序和前序，中序和后序）

来源：互联网发布：uv胶多少钱淘宝编辑：程序博客网时间：2024/06/04 14:04

1.根据中序和前序

原题目

public class Solution {    public TreeNode buildTree(int[] preorder, int[] inorder) {        if (preorder == null || preorder.length == 0 || inorder == null || inorder.length == 0) {            return null;        }        return build(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);    }    public TreeNode build(int[] preorder, int leftP, int rightP, int[] inorder, int leftI, int rightI) {        if (leftP > rightP || leftI > rightI) {            return null;        }        TreeNode root = new TreeNode(preorder[leftP]);        int indexInInOrder = findRootIndexInOrder(inorder, leftI, rightI, root.val);        int offset = indexInInOrder - leftI - 1;        TreeNode left = build(preorder, leftP + 1, leftP + 1 + offset, inorder, leftI, indexInInOrder - 1);        TreeNode right = build(preorder, leftP + offset + 2, rightP, inorder, indexInInOrder + 1, rightI);        root.left = left;        root.right = right;        return root;    }    public int findRootIndexInOrder(int[] inorder, int left, int right, int target) {        int result = 0;        for (int i = left; i <= right; i++) {            if (inorder[i] == target) {                result = i;                break;            }        }        return result;    }}

offset很重要，一步到位

2.根据中序和后序

public class Solution {    public TreeNode buildTree(int[] inorder, int[] postorder) {        if (inorder == null || inorder.length == 0 || postorder == null || postorder.length == 0) {            return null;        }        return build(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);    }    public TreeNode build(int[] inorder, int leftI, int rightI, int[] postorder, int leftP, int rightP) {        if (leftI > rightI || leftP > rightP) {            return null;        }        TreeNode root = new TreeNode(postorder[rightP]);        int indexInInOrder = findRootIndexInOrder(inorder, leftI, rightI, root.val);        int offset = indexInInOrder - 1 - leftI;        TreeNode left = build(inorder, leftI, indexInInOrder - 1, postorder, leftP, leftP + offset);        TreeNode right = build(inorder, indexInInOrder + 1, rightI, postorder, leftP + offset + 1, rightP - 1);        root.left = left;        root.right = right;        return root;    }    public int findRootIndexInOrder(int[] inorder, int left, int right, int target) {        int result = 0;        for (int i = left; i <= right; i++) {            if (inorder[i] == target) {                result = i;                break;            }        }        return result;    }}

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