1. 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 between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Note:

  • All of the nodes' values will be unique.
  • p and q are different and both values will exist in the binary tree.

对于root节点,如果左右子树分别有一个节点,则root是一个公共祖先

解1 递归

class Solution {
public:
TreeNode *ans;
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
LCA(root, p, q);
return ans;
}
bool LCA(TreeNode *root, TreeNode *p, TreeNode *q){
if(root == NULL)return false;
int l = LCA(root->left, p, q) ? 1 : 0; // p或者q在左子树
int r = LCA(root->right, p, q) ? 1 : 0; // p或者q在右子树
int mid = (root == p || root == q) ? 1 : 0; //找到了p或者q
if(mid + l + r >= 2)ans = root; // =2时,左右各一个;>2时,一个是另一个的先驱节点
return mid + l + r > 0; // >0表明找到了p或者q
}
};

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