根据一棵树的中序遍历与后序遍历构造二叉树。
注意: 你可以假设树中没有重复的元素。
例如,给出
中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]
返回如下的二叉树:
3
/ \
9 20
/ \
15 7
思路:与leetcode105(https://blog.csdn.net/weixin_38823568/article/details/82982621)
Java代码:
class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {
return dfs(inorder, 0, inorder.length, postorder, postorder.length-1);
}
private TreeNode dfs(int[] inorder, int is, int ie, int[] postorder, int pe){
if(is == ie)
return null;
if(1 == ie-is)
return new TreeNode(inorder[is]);
TreeNode root = new TreeNode(postorder[pe]);
int index = is;
for(; index < ie && inorder[index] != postorder[pe]; index++);
root.left = dfs(inorder, is, index, postorder, pe-(ie-index));
root.right = dfs(inorder, index+1, ie, postorder, pe-1);
return root;
}
}