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leetcode 236: Lowest Common Ancestor of a Binary Tree
Lowest Common Ancestor of a Binary Tree
Total Accepted: 1628 Total Submissions: 5660Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allowa node to be a descendant of itself).”
_______3______ / ___5__ ___1__ / \ / 6 _2 0 8 / 7 4
For example, the lowest common ancestor (LCA) of nodes 5
and
1
is 3
. Another example is LCA of nodes 5
and
4
is 5
, since a node can be a descendant of itself according to the LCA definition.
[思路]
假设是普通二叉树, 而不是BST. 则应该遍历节点, 先找到p,q. 同一时候记录下从root到该几点的路径. 之后比較路径,最后一个同样的节点便是LCA.
[CODE]
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { //2, 1 public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if(root==null || p==null || q==null) return null; List<TreeNode> pathp = new ArrayList<>(); List<TreeNode> pathq = new ArrayList<>(); pathp.add(root); pathq.add(root); getPath(root, p, pathp); getPath(root, q, pathq); TreeNode lca = null; for(int i=0; i<pathp.size() && i<pathq.size(); i++) { if(pathp.get(i) == pathq.get(i)) lca = pathp.get(i); else break; } return lca; } private boolean getPath(TreeNode root, TreeNode n, List<TreeNode> path) { if(root==n) { return true; } if(root.left!=null) { path.add(root.left); if(getPath(root.left, n, path)) return true; path.remove(path.size()-1); } if(root.right!=null) { path.add(root.right); if(getPath(root.right, n, path)) return true; path.remove(path.size()-1); } return false; } }
leetcode 236: Lowest Common Ancestor of a Binary Tree
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