105. Construct Binary Tree from Preorder and Inorder Traversal
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2022-05-18 19:35:40
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Given preorder and inorder traversal of a tree, construct the binary tree.
Solution1: Recursive做法
Example:
1
2 3
4 5 6 7
8 9 10
In order: 8 4 9 2 5 1* 10 6 3 7
Preorder: 1* 2 4 8 9 5 3 6 10 7
与106题 http://www.jianshu.com/p/bacca777b12b 类似
思路:
Preorder的第一位是根节点,顺序:根节点,左树,右树;Inorder顺序:左树,根节点,右树。
利用此特性,每次在每一段pretorder序列中,得到第一位元素,在inorder序列中查找此元素位置pos,inorder序列中pos的前、后部分a(8 4 9 2 5)、b(10 6 3 7)分别作为下次递归的inorder序列input;
并利用a、b部分的长度/位置在pretorder序列中找到相应部分c(2 4 8 9 5)、d(3 6 10 7),作为下次递归的postorder序列input。
重复此过程(Top-down Recursion)。
在inorder查找的过程利用hashmap,故
Time Complexity: O(N) Space Complexity: O(N)
Solution2: Iterative做法
思路:
Solution1 Code:
class Solution1 {
private HashMap<Integer, Integer> inorder_map = new HashMap<Integer, Integer>();
public TreeNode buildTree(int[] preorder, int[] inorder) {
if (inorder == null || preorder == null || inorder.length != preorder.length)
return null;
// build hashmap from Inorder array
for (int i = 0; i < inorder.length; ++i) {
inorder_map.put(inorder[i], i);
}
// recursively build the tree
return build(inorder, 0, inorder.length - 1, preorder, 0, preorder.length - 1);
}
public TreeNode build(int[] inorder, int in_start, int in_end, int[] preorder, int pre_start, int pre_end) {
if(in_start > in_end || pre_start > pre_end)
return null;
int root_value = preorder[pre_start];
TreeNode root = new TreeNode(root_value);
int pos_inorder = inorder_map.get(root_value);
int pre_left_length = pos_inorder - in_start;
root.left = build(inorder, in_start, pos_inorder - 1, preorder, pre_start + 1, pre_start + pre_left_length);
root.right = build(inorder, pos_inorder + 1, in_end, preorder, pre_start + pre_left_length + 1, pre_end);
return root;
}
}
Solution1.Round1 Code:
class Solution {
private Map<Integer, Integer> in_map;
public TreeNode buildTree(int[] preorder, int[] inorder) {
in_map = new HashMap<>();
for(int i = 0; i < inorder.length; i++) {
in_map.put(inorder[i], i);
}
return helper(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
}
private TreeNode helper(int[] preorder, int pre_start, int pre_end, int[] inorder, int in_start, int in_end) {
if(pre_start > pre_end || in_start > in_end) {
return null;
}
int root_val = preorder[pre_start];
int root_in_index = in_map.get(root_val);
TreeNode node = new TreeNode(root_val);
int relat_root_in_index = root_in_index - in_start;
node.left = helper(preorder, pre_start + 1, pre_start + 1 + relat_root_in_index - 1, inorder, in_start, root_in_index - 1);
node.right = helper(preorder, pre_start + 1 + relat_root_in_index, pre_end, inorder, root_in_index + 1, in_end);
return node;
}
}
change End to 元素边界
class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
if(preorder == null || inorder == null || preorder.length == 0 || inorder.length == 0) return null;
Map<Integer, Integer> in_map = new HashMap<>();
for(int i = 0; i < inorder.length; i++) {
in_map.put(inorder[i], i);
}
return dfsBuild(in_map, preorder, 0, preorder.length, inorder, 0, inorder.length);
}
private TreeNode dfsBuild(Map<Integer, Integer> in_map, int[] preorder, int pre_s, int pre_e, int[] inorder, int in_s, int in_e) {
if(pre_s == pre_e || in_s == in_e) return null;
int cur_val = preorder[pre_s];
int in_index = in_map.get(cur_val);
int rela_in_index = in_index - in_s;
TreeNode root = new TreeNode(cur_val);
root.left = dfsBuild(in_map, preorder, pre_s + 1, pre_s + 1 + rela_in_index, inorder, in_s, in_index);
root.right = dfsBuild(in_map, preorder, pre_s + 1 + rela_in_index, pre_e, inorder, in_index + 1, in_e);
return root;
}
}
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