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106. Construct Binary Tree from Inorder and Postorder Traversal
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
inorder = [9,3,15,20,7] postorder = [9,15,7,20,3]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
参考资料
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/**
* Runtime: 0 ms, faster than 100.00% of Java online submissions for Construct Binary Tree from Inorder and Postorder Traversal.
* Memory Usage: 41.3 MB, less than 14.54% of Java online submissions for Construct Binary Tree from Inorder and Postorder Traversal.
*/
public TreeNode buildTree(int[] inorder, int[] postorder) {
return buildTree(inorder, postorder, 0, inorder.length - 1, postorder.length - 1);
}
private TreeNode buildTree(int[] inorder, int[] postorder, int instart, int inend, int postend) {
if (postend < 0 || instart > inend) {
return null;
}
int val = postorder[postend];
int index = inend;
for (; index >= instart; index--) {
if (inorder[index] == val) {
break;
}
}
TreeNode result = new TreeNode(val);
result.left = buildTree(inorder, postorder, instart, index - 1, postend - (inend - index) - 1);
result.right = buildTree(inorder, postorder, index + 1, inend, postend - 1);
return result;
}
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/**
*在自己实现的基础上，参考： https://leetcode.cn/problems/construct-binary-tree-from-inorder-and-postorder-traversal/solutions/426738/cong-zhong-xu-yu-hou-xu-bian-li-xu-lie-gou-zao-14/comments/2236345 解决了下标错误的问题。
*/
public TreeNode buildTree(int[] inorder, int[] postorder) {
if (inorder == null || postorder == null || inorder.length != postorder.length) {
return null;
}
Map<Integer, Integer> numToIndex = new HashMap<>();
for (int i = 0; i < inorder.length; i++) {
numToIndex.put(inorder[i], i);
}
return buildTree(inorder, postorder, numToIndex, 0, inorder.length - 1, 0, postorder.length - 1);
}
private TreeNode buildTree(int[] inorder, int[] postorder, Map<Integer, Integer> numToIndex, int inStart, int inEnd, int postStart, int postEnd) {
if (inStart > inEnd || postStart > postEnd) {
return null;
}
int value = postorder[postEnd];
TreeNode root = new TreeNode(value);
int index = numToIndex.get(value);
int size = index - inStart;
root.left = buildTree(inorder, postorder, numToIndex, inStart, index - 1, postStart, postStart + size - 1);
root.right = buildTree(inorder, postorder, numToIndex, index + 1, inEnd, postStart + size, postEnd - 1);
return root;
}
