Check on some random trees what are the depths of leaves merging in optimal answer.

Lets precalc two arrays for each of trees: leavesOdd and leavesEven (leaves of odd and even depths from the root of tree, it doesnt matter what to choose as a root).
Then you can notice that answer is always either 2 or 3.

2 can be achieved only when (leavesOdd₀ = leavesOdd₁ and leavesEven₀ = leavesEven₁) or (leavesOdd₀ = leavesEven₁ and leavesEven₀ == leavesOdd₁). It's because all the paths will be of the same parity and thus can be colored in 2 colors (color the first root into 1 and then layer by layer alterate color).

The other cases it's 3 because in any path which is of nonoptimal parity you can color leaf in the third color and connect it to path of whatever parity tou need.

4-dimensional dp for max amount of certain type of characters.

Let's calc dp[i][j][k][l] — max amount of '|' characters we can collect on our path from (0, 0) to (i, j) while having k '<' characters collected and l '-' characters collected. dp array should be initialized with negative infinite values. Transitions are dependant on character a[i][j]. So it's like:
1. (a[i][j] == '>')
dp[i][j][k][l] = max(dp[i][j][k][l], dp[i - 1][j][k - 1][l], dp[i][j - 1][k - 1][l])
2. (a[i][j] == '-')
dp[i][j][k][l] = max(dp[i][j][k][l], dp[i - 1][j][k][l - 1], dp[i][j - 1][k][l - 1])
3. (a[i][j] == '|')
dp[i][j][k][l] = max(dp[i][j][k][l], dp[i - 1][j][k][l] + 1, dp[i][j - 1][k][l] + 1)

The answer is

And you also need to optimize it from O(N⁴) memory to O(N³). Just store only previous and current layers of dp.