Given 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 p and q as the lowest node in T that has both p and q…

Given 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 p and q as the lowest node in T that has both p and q…

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/ Given 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…

Given 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…

Given 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…

  Given 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 p and q as the lowest node in T that has both p and…

Given 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 p and q as the lowest node in T that has both p and q…

解题思路一: DFS到每个节点的路径,根据路径算出LCA: public class Solution { public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { if (root == p || root == q) return root; List pList = new ArrayList<TreeNode>(), qList = new ArrayList<TreeNode>…

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: TreeNode* lowestCommonAncestor(TreeNode* root, TreeNo…

如果一个节点的左右子树上分别有两个节点,那么这棵树是祖先,但是不一定是最小的,但是从下边开始判断,找到后一直返回到上边就是最小的. 如果一个节点的左右子树上只有一个子树上遍历到了节点,那么那个子树可能是一个节点的祖先,也可能是两个节点的祖先,如果是一个节点的祖先,那么公共祖先还在上边,还需要返回结果进行判断,如果是两个节点的祖先,那么最小公共祖先就是这个或者在下边,总之,返回有结果的那个子树就是了. 所以思路就是,往下遍历,直到找到节点然后返回,返回以后判断此时左右子树的情况,根据情况返回根节点…