题目：
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
preorder = [3,9,20,15,7] inorder = [9,3,15,20,7]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
实现：
- 递归方式，根据二叉树中序遍历和前续遍历，创建二叉树。中序遍历： 左-根-右，前续遍历： 根-左-右， 所以可以通过前续遍历的第一个节点找到根节点，并生成根节点， 因为二叉树没有重复节点， 就 可以在中序遍历数据中找到根节点，再分成左右两个子树，递归执行下去，遍历完所有节点。 通过preorder获取根节点， 通过inorder生成左右子树，查找节点在中序遍历数据组里位置，java可以通过HashMap实现。
- 不使用map情况下 时间复杂度 O(N^2) , 使用map查找O(N)
/** java
* 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 {
public TreeNode buildTree(int[] preorder, int[] inorder) {
Map<Integer, Integer> treemap = new HashMap <>();
for (int i=0; i<inorder.length; i++){
treemap.put(inorder[i], i);
}
AtomicInteger rootindex = new AtomicInteger(0);
return helper(preorder, inorder, 0, inorder.length -1, rootindex, treemap);
}
public TreeNode helper(int[] preorder, int[] inoder, int left, int right, AtomicInteger index, Map<Integer, Integer> treemap) {
if (left > right){
return null;
}
//new node, set root
int val = preorder[index.intValue()];
TreeNode node = new TreeNode(val);
index
//search inorder
int root = treemap.get(val);
//left substree
node.left = helper(preorder, inoder, left, root-1, index, treemap);
//right substree
node.right = helper(preorder, inoder, root+1, right, index, treemap);
return node ;
}
}参考及引用:
techiedelight-construct-binary-tree-from-inorder-preorder-traversal
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