105. 从前序与中序遍历序列构造二叉树
作者:互联网
给定一棵树的前序遍历 preorder 与中序遍历 inorder。请构造二叉树并返回其根节点。
来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
import java.util.HashMap;
import java.util.Map;
class Solution {
private static TreeNode build(int[] preorder, int preStart, int preEnd, Map<Integer, Integer> inMap, int inStart, int inEnd) {
if (preStart == preEnd) {
return new TreeNode(preorder[preStart]);
}
if (preStart > preEnd) {
return null;
}
int inVal = preorder[preStart];
TreeNode root = new TreeNode(inVal);
int inPos = inMap.get(inVal);
int leftLength = inPos - inStart;
int rightLength = inEnd - inPos;
root.left = build(preorder, preStart + 1, preStart + leftLength, inMap, inStart, inPos - 1);
root.right = build(preorder, preEnd - rightLength + 1, preEnd, inMap, inPos + 1, inEnd);
return root;
}
public static TreeNode buildTree(int[] preorder, int[] inorder) {
if (preorder == null || preorder.length == 0 || inorder == null || inorder.length == 0 || preorder.length != inorder.length) {
return null;
}
Map<Integer, Integer> inMap = new HashMap<>();
for (int i = 0; i < inorder.length; ++i) {
inMap.put(inorder[i], i);
}
return build(preorder, 0, preorder.length - 1, inMap, 0, inorder.length - 1);
}
public static void main(String[] args) {
int[] pre = {3, 9, 20, 15, 7};
int[] in = {9, 3, 15, 20, 7};
buildTree(pre, in);
}
}
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;
}
}
标签:preorder,TreeNode,preStart,int,中序,inMap,二叉树,inorder,105 来源: https://www.cnblogs.com/tianyiya/p/15652850.html