106 construct tree with inorder

2017-10-24  本文已影响0人  Fei_JOB
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder == null || postorder == null) return null;
        if(inorder.length != postorder.length) return null;
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for(int i = 0; i < inorder.length; i++){
            map.put(inorder[i], i);
        }
        int len = inorder.length;
        return helper(inorder, 0, len-1, postorder, 0, len -1, map);
    }
    public TreeNode helper(int[] inorder, int s1, int e1, int[] postorder, int s2, int e2, Map<Integer, Integer> map){
        if(s2 > e2) return null;
        TreeNode node = new TreeNode(postorder[e2]);
        if(s2 == e2)  return node;
        int inPos = map.get(postorder[e2]);
        int numLeft = inPos - s1;
        node.left = helper(inorder, s1, inPos-1, postorder, s2, s2+ numLeft -1, map);
        node.right = helper(inorder, inPos + 1, e1, postorder, s2+numLeft, e2-1, map);
        return node;
    }
}
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